Add Maths · 8 Coordinate geometry of the circle

8 Coordinate geometry of the circle

Before you start

The distance between two points, $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ , and the midpoint of a line segment.
Gradients: the gradient of a line through two points, and that perpendicular gradients multiply to $- 1$ , so $\frac{4}{3}$ becomes $- \frac{3}{4}$ .
The equation of a line through $(x_{1}, y_{1})$ with gradient $m$ : $y - y_{1} = m (x - x_{1})$ .
Completing the square, e.g. $x^{2} - 6 x = {(x - 3)}^{2} - 9$ . It is not on the List of formulas, so you must recall it.
Substituting a linear equation into a quadratic one and solving, and the discriminant conditions $b^{2} - 4 a c > 0$ , $= 0$ , $< 0$ , which you must also recall.
Simplifying surds, e.g. $\sqrt{20} = 2 \sqrt{5}$ .

Syllabus objectives covered (4)

Know and use the equation of a circle with radius r and centre (a, b).
Solve problems involving the intersection of a circle and a straight line.
Solve problems involving tangents to a circle.
Solve problems involving the intersection of two circles.

Part 1 of 7

Where the equation of a circle comes from

A circle is every point that is the same distance $r$ , the radius, from one fixed point, the centre. Everything in this topic comes back to that sentence: distances from the centre, and the right angles they make.
Put the centre at $C (a, b)$ and let $P (x, y)$ be any point on the circle. The horizontal distance from $C$ to $P$ is $x - a$ and the vertical distance is $y - b$ . These are the two shorter sides of a right-angled triangle whose hypotenuse is the radius, so Pythagoras gives the equation.
${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$
graph diagram — not drawn yet
Square-scaled axes for −5 ≤ x ≤ 7 and −4 ≤ y ≤ 8 showing the circle (x − 1)² + (y − 2)² = 25, with the centre C(1, 2) marked with a dot and labelled, and the point P(4, 6) marked on the circle and labelled. Draw the radius CP as a solid segment labelled 'r = 5', a dashed horizontal segment from C to (4, 2) labelled 'x − a = 3' and a dashed vertical segment from (4, 2) to P labelled 'y − b = 4', with a right-angle mark at (4, 2).
The equation is Pythagoras for this triangle: $3^{2} + 4^{2} = 5^{2}$ , so $P (4, 6)$ lies on the circle.
This equation is on the List of formulas, so you do not have to memorise it. What you do need is to read it correctly, and there are two traps in reading it.
Trap one: the signs. The equation subtracts the centre's coordinates, so ${(x + 3)}^{2} + {(y - 2)}^{2} = 16$ is really ${(x - (- 3))}^{2} + {(y - 2)}^{2} = 4^{2}$ . The centre is $(- 3, 2)$ , not $(3, - 2)$ . Change the sign of the number inside each bracket.
Trap two: the right-hand side is $r^{2}$ , not $r$ . Here $r^{2} = 16$ , so the radius is $4$ . If the right-hand side is $18$ , the radius is $\sqrt{18} = 3 \sqrt{2}$ ; on the non-calculator paper, leave it as a simplified surd.
To write the equation of a circle you need only the centre and $r^{2}$ . For the circle with centre $(1, 2)$ through the point $(4, 6)$ , $r^{2} = {(4 - 1)}^{2} + {(6 - 2)}^{2} = 25$ , so the equation is ${(x - 1)}^{2} + {(y - 2)}^{2} = 25$ . You never needed $r$ itself.
If you are given the two ends of a diameter, the centre is their midpoint, and the radius is half the distance between them.
To decide whether a point is inside, on or outside a circle, substitute it into the left-hand side ${(x - a)}^{2} + {(y - b)}^{2}$ . That is the squared distance from the centre. Less than $r^{2}$ means inside, equal to $r^{2}$ means on the circle, greater than $r^{2}$ means outside.

Part 2 of 7

Centre and radius from the expanded form

Multiply out ${(x - 1)}^{2} + {(y - 2)}^{2} = 25$ and collect everything on one side. It is the same circle, written so that the centre is hidden.
$x^{2} + y^{2} - 2 x - 4 y - 20 = 0$
Two features mark an equation like this as a circle: the $x^{2}$ and $y^{2}$ terms have the same coefficient, and there is no $x y$ term. If either fails, the curve is not a circle.
To get the centre back, complete the square on the $x$ terms and on the $y$ terms separately. For $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ , group the letters and move the constant to the right.
$(x^{2} - 6 x) + (y^{2} + 4 y) = 12$
Halve the coefficient of $x$ to get $- 3$ , so $x^{2} - 6 x = {(x - 3)}^{2} - 9$ . Halve the coefficient of $y$ to get $2$ , so $y^{2} + 4 y = {(y + 2)}^{2} - 4$ . The $- 9$ is there because ${(x - 3)}^{2}$ expands to $x^{2} - 6 x + 9$ , which is $9$ more than you started with; the $- 4$ is there for the same reason.
${(x - 3)}^{2} - 9 + {(y + 2)}^{2} - 4 = 12$
Move the constants to the right: ${(x - 3)}^{2} + {(y + 2)}^{2} = 25$ . The centre is $(3, - 2)$ and the radius is $5$ . The trap is dropping the $- 9$ and $- 4$ : that leaves $r^{2} = 12$ , a wrong radius with a correct-looking centre.
Doing the same steps with letters gives a shortcut you can use as a check. For $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , completing the square gives ${(x + g)}^{2} + {(y + f)}^{2} = g^{2} + f^{2} - c$ .
$centre (- g, - f), r = \sqrt{g^{2} + f^{2} - c}$
Check it on the example: $2 g = - 6$ gives $g = - 3$ , $2 f = 4$ gives $f = 2$ , and $c = - 12$ . So the centre is $(3, - 2)$ and $r^{2} = 9 + 4 + 12 = 25$ . Notice that $c$ is subtracted, so a negative $c$ makes $r^{2}$ larger.
If $x^{2}$ and $y^{2}$ have a coefficient other than $1$ , divide the whole equation by it first. $3 x^{2} + 3 y^{2} - 12 x + 6 y = 0$ becomes $x^{2} + y^{2} - 4 x + 2 y = 0$ , so the centre is $(2, - 1)$ and $r^{2} = 4 + 1 - 0 = 5$ , giving $r = \sqrt{5}$ .
Completing the square can also show that an equation is not a circle at all. $x^{2} + y^{2} - 2 x + 4 y + 10 = 0$ becomes ${(x - 1)}^{2} + {(y + 2)}^{2} = - 5$ . A sum of two squares cannot be negative, so no point fits. For a circle, $r^{2}$ must come out positive.

Part 3 of 7

Where a line meets a circle

This is the line-and-curve method from simultaneous equations. Make $y$ (or $x$ ) the subject of the line, substitute it into the circle's equation with brackets, and you get a quadratic in one letter.
That quadratic decides how the line and circle meet. Two different roots: the line cuts the circle at two points, and the part of the line between them is a chord. A repeated root: one point, so the line is a tangent. No real roots: the line does not meet the circle.
$b^{2} - 4 a c > 0 (two points), b^{2} - 4 a c = 0 (tangent), b^{2} - 4 a c < 0 (no intersection)$
Take the circle ${(x - 2)}^{2} + {(y - 1)}^{2} = 5$ and the line $y = 2 x - 3$ . Substitute: ${(x - 2)}^{2} + {(2 x - 4)}^{2} = 5$ . Expanding gives $5 x^{2} - 20 x + 20 = 5$ , so $5 x^{2} - 20 x + 15 = 0$ . Divide by $5$ .
$x^{2} - 4 x + 3 = 0$
Here $b^{2} - 4 a c = 16 - 12 = 4 > 0$ , so there are two points. Factorise: $(x - 1) (x - 3) = 0$ . Find each $y$ from the line, not the circle, so that each $x$ gets exactly one partner: $x = 1$ gives $y = - 1$ and $x = 3$ gives $y = 3$ . The points are $(1, - 1)$ and $(3, 3)$ .
Keep the circle and move the line. $y = 2 x + 2$ gives $5 x^{2} = 0$ , a repeated root, so this line is a tangent touching at $(0, 2)$ . $y = 2 x + 5$ gives $5 x^{2} + 12 x + 15 = 0$ , where $b^{2} - 4 a c = 144 - 300 = - 156 < 0$ , so this line misses the circle.
graph diagram — not drawn yet
Square-scaled axes for −3 ≤ x ≤ 6 and −4 ≤ y ≤ 6 showing the circle (x − 2)² + (y − 1)² = 5 (centre (2, 1) marked, radius √5 ≈ 2.24) and three parallel lines of gradient 2: y = 2x − 3 crossing the circle at (1, −1) and (3, 3), y = 2x + 2 touching it at (0, 2), and y = 2x + 5 passing clear of it. Mark the intersection and contact points with dots, and label each line with its equation and 'b² − 4ac > 0', 'b² − 4ac = 0' or 'b² − 4ac < 0'.
Same gradient, different intercepts: as the line moves away from the centre, $b^{2} - 4 a c$ goes from positive to zero to negative.
When a question asks you to Determine whether a line is a tangent, cuts the circle or misses it, the value of $b^{2} - 4 a c$ is your evidence, and the conclusion must be written in words. The number on its own does not answer the question.
The length of a chord is the distance between its two end points. For $(1, - 1)$ and $(3, 3)$ it is $\sqrt{2^{2} + 4^{2}} = \sqrt{20} = 2 \sqrt{5}$ . That equals the diameter, $2 \sqrt{5}$ , which makes sense: $y = 2 x - 3$ passes through the centre $(2, 1)$ .
One more fact about chords. A circle is symmetrical about every line through its centre, so the perpendicular bisector of any chord passes through the centre. Two points on a circle therefore give you a line that the centre must lie on.
Example: a circle passes through $A (2, 0)$ and $B (0, 4)$ , and its centre lies on $y = x$ . The midpoint of $A B$ is $(1, 2)$ and $A B$ has gradient $- 2$ , so the perpendicular bisector has gradient $\frac{1}{2}$ : $y - 2 = \frac{1}{2} (x - 1)$ . Put $y = x$ into it: $2 x - 4 = x - 1$ , so $x = 3$ . The centre is $(3, 3)$ and $r^{2} = {(3 - 2)}^{2} + {(3 - 0)}^{2} = 10$ .

Part 4 of 7

The tangent at a point on the circle

A tangent touches the circle at one point, the point of contact. At that point the tangent is perpendicular to the radius. That fact turns a tangent question into a gradient question, and no calculus is needed.
The method has three steps. Find the gradient of the radius from the centre $C$ to the point of contact $P$ . Take the negative reciprocal to get the tangent's gradient. Write the equation of the line through $P$ with that gradient.
Take ${(x - 1)}^{2} + {(y - 2)}^{2} = 25$ and $P (4, 6)$ . First check that $P$ is on the circle: $3^{2} + 4^{2} = 25$ . The gradient of $C P$ is $\frac{6 - 2}{4 - 1} = \frac{4}{3}$ , so the tangent's gradient is $- \frac{3}{4}$ .
Use $P$ in the line equation, because $P$ is the point the tangent passes through. The centre is not on the tangent at all.
$y - 6 = - \frac{3}{4} (x - 4) \Rightarrow 4 y - 24 = - 3 x + 12 \Rightarrow 3 x + 4 y = 36$
graph diagram — not drawn yet
Square-scaled axes for −5 ≤ x ≤ 13 and −4 ≤ y ≤ 10 showing the circle (x − 1)² + (y − 2)² = 25 with the centre C(1, 2) and the point P(4, 6) marked and labelled. Draw the radius CP and the tangent 3x + 4y = 36 through P, crossing the axes at (12, 0) and (0, 9), with a right-angle mark between CP and the tangent at P. Label CP 'gradient 4/3' and the tangent 'gradient −3/4'.
The tangent is perpendicular to the radius at $P$ , so its gradient is the negative reciprocal of $\frac{4}{3}$ .
Check the line by putting $P$ into it: $3 (4) + 4 (6) = 12 + 24 = 36$ . If $P$ does not satisfy your tangent, something has gone wrong before the last line.
The awkward case is a horizontal or vertical radius. On the same circle, $(6, 2)$ is level with the centre, so the radius is horizontal and the tangent there is the vertical line $x = 6$ . At $(1, 7)$ the radius is vertical and the tangent is $y = 7$ . There is no negative reciprocal of $0$ to calculate; picture the radius instead.
Tangents also make right-angled triangles. If a tangent from an outside point $Q$ touches the circle at $T$ , the angle $C T Q$ is $90^{\circ}$ , so Pythagoras gives the length of the tangent.
$Q T^{2} = C Q^{2} - r^{2}$
For $Q (13, 7)$ and the same circle, $C Q^{2} = 12^{2} + 5^{2} = 169$ , so $Q T^{2} = 169 - 25 = 144$ and the tangent has length $12$ . You never need the coordinates of $T$ .

Part 5 of 7

When the line has an unknown: tangent means a repeated root

Sometimes you are not given the point of contact. The question gives a line such as $y = x + k$ and asks for the values of $k$ that make it a tangent. You cannot use the gradient of a radius, because you do not know where the radius ends.
Use the intersection method instead. Substitute the line into the circle to get a quadratic in $x$ . A tangent meets the circle at exactly one point, so that quadratic must have a repeated root.
$b^{2} - 4 a c = 0$
The coefficients $a$ , $b$ and $c$ now contain $k$ , so this condition is an equation in $k$ . When it is a quadratic it gives two values of $k$ : two parallel tangents, one on each side of the circle. That is what you should expect, because a circle has two tangents in every direction. The worked examples finish one of these.
The unknown can be in the circle instead of the line. A circle with centre $(0, 0)$ touches the line $x + y = 4$ . Substitute $y = 4 - x$ into $x^{2} + y^{2} = r^{2}$ and collect terms.
$2 x^{2} - 8 x + 16 - r^{2} = 0$
Set $b^{2} - 4 a c = 0$ : $64 - 8 (16 - r^{2}) = 0$ , so $8 r^{2} = 64$ and $r^{2} = 8$ . The circle is $x^{2} + y^{2} = 8$ , and the repeated root $x = 2$ gives the point of contact $(2, 2)$ .
A question can also ask when the line meets the circle at two points, or not at all. Then use $b^{2} - 4 a c > 0$ or $b^{2} - 4 a c < 0$ and solve the quadratic inequality. For $x^{2} + y^{2} = 8$ and $x + y = c$ , the discriminant is $64 - 4 c^{2}$ , so the line meets the circle at two points when $- 4 < c < 4$ , touches it when $c = \pm 4$ , and misses it when $c < - 4$ or $c > 4$ .
To show that a given line is a tangent, do the same substitution and show that the quadratic has a repeated root, either with $b^{2} - 4 a c = 0$ or by writing it as a perfect square. Then write the conclusion: repeated root, so the line is a tangent.

Part 6 of 7

Two circles: intersect, touch or do not intersect

For two circles, three numbers decide everything: the distance $d$ between the centres, and the two radii $r_{1}$ and $r_{2}$ . Find both centres and both radii first, then find $d$ with the distance formula.
Start with circles outside each other. If $d > r_{1} + r_{2}$ , the radii cannot reach across the gap, so the circles do not intersect. If $d = r_{1} + r_{2}$ , they just reach, and the circles touch at one point. If $d < r_{1} + r_{2}$ , they can overlap and cross at two points.
$d > r_{1} + r_{2} (apart), d = r_{1} + r_{2} (touch), d < r_{1} + r_{2} (may cross)$
The awkward case is one circle inside the other, so also compare $d$ with the difference $| r_{1} - r_{2} |$ . If $d = | r_{1} - r_{2} |$ , the smaller circle touches the larger one from inside. If $d < | r_{1} - r_{2} |$ , the smaller circle sits inside the larger one without meeting it. Two points of intersection need both conditions together.
$| r_{1} - r_{2} | < d < r_{1} + r_{2}$
graph diagram — not drawn yet
Five small panels in a row, no axes. Each shows two circles with centres C₁ and C₂ marked and the segment C₁C₂ drawn and labelled d: (1) two circles apart, labelled 'd > r₁ + r₂: no intersection'; (2) two circles touching from outside at one marked point on C₁C₂, labelled 'd = r₁ + r₂: touch'; (3) two overlapping circles with both crossing points marked, labelled '|r₁ − r₂| < d < r₁ + r₂: two points'; (4) a small circle inside a large one, touching it at one marked point on the line C₁C₂, labelled 'd = |r₁ − r₂|: touch'; (5) a small circle wholly inside a large one without touching, labelled 'd < |r₁ − r₂|: no intersection'.
Only the middle panel has two points of intersection, and it needs both inequalities.
Example: ${(x - 1)}^{2} + {(y + 2)}^{2} = 9$ and ${(x - 5)}^{2} + {(y - 1)}^{2} = 4$ . The centres are $(1, - 2)$ and $(5, 1)$ , so $d = \sqrt{4^{2} + 3^{2}} = 5$ . The radii are $3$ and $2$ , and $3 + 2 = 5 = d$ , so the circles touch from outside.
The point of contact lies on the line joining the centres, a distance $r_{1} = 3$ from the first centre. That is $\frac{3}{5}$ of the way from $(1, - 2)$ to $(5, 1)$ .
$(1 + \frac{3}{5} (4), - 2 + \frac{3}{5} (3)) = (\frac{17}{5}, - \frac{1}{5})$
An inside example: $x^{2} + y^{2} = 36$ and ${(x - 3)}^{2} + {(y - 4)}^{2} = 1$ . Here $d = \sqrt{9 + 16} = 5$ and $| 6 - 1 | = 5 = d$ , so the small circle touches the large one from inside. The contact point is $6$ units from the origin in the direction of $(3, 4)$ : $\frac{6}{5} (3, 4) = (\frac{18}{5}, \frac{24}{5})$ .
With surd radii, simplify before you add, then compare squares so you need no decimals. Radii $\sqrt{20}$ and $\sqrt{5}$ are $2 \sqrt{5}$ and $\sqrt{5}$ , so $r_{1} + r_{2} = 3 \sqrt{5}$ and ${(r_{1} + r_{2})}^{2} = 45$ . Compare $d^{2}$ with $45$ .

Part 7 of 7

Where two circles cross: the common chord

To find where two circles meet, write both in expanded form, with $x^{2}$ and $y^{2}$ having coefficient $1$ and $0$ on the right-hand side. Then subtract one equation from the other. The $x^{2}$ and $y^{2}$ terms cancel, and a linear equation is left.
Why that line matters: each point where the circles meet satisfies both equations, so it also satisfies their difference. The linear equation is therefore the line through both points of intersection, the common chord.
Substitute that line into either circle to get a quadratic. Solve it and pair each $x$ with its $y$ from the line, exactly as in the line-and-circle method. The simpler circle makes the algebra shorter.
The quadratic also tells you how the circles sit. Two roots: they intersect at two points. A repeated root: they touch, and the line is the tangent to both circles at the point of contact. No real roots: they do not intersect.
Check this on the touching pair from the previous section. Expanded, they are $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$ and $x^{2} + y^{2} - 10 x - 2 y + 22 = 0$ . Subtracting gives $8 x + 6 y - 26 = 0$ , so $4 x + 3 y = 13$ . Substituting into the first circle leads to ${(5 x - 17)}^{2} = 0$ : a repeated root at $x = \frac{17}{5}$ , the same point of contact as before.
Two traps. If one equation starts $2 x^{2} + 2 y^{2}$ , divide it by $2$ before subtracting, or the squares do not cancel. If both circles have the same centre, the subtraction leaves only numbers, such as $5 = 0$ : circles with the same centre and different radii never meet.
A quick check on your chord: it is perpendicular to the line joining the centres. In the touching example the centres give gradient $\frac{3}{4}$ and $4 x + 3 y = 13$ has gradient $- \frac{4}{3}$ , and $\frac{3}{4} \times (- \frac{4}{3}) = - 1$ .

Worked examples

Example 1. The circle $C$ has equation $2 x^{2} + 2 y^{2} - 8 x + 12 y - 6 = 0$ . Find the centre and radius of $C$ , and determine whether each of the points $(5, 0)$ and $(4, - 1)$ lies inside, on or outside $C$ .

Show the working

The coefficients of $x^{2}$ and $y^{2}$ are $2$ . Divide the whole equation by $2$ first, so that completing the square works.
$x^{2} + y^{2} - 4 x + 6 y - 3 = 0$
Group the $x$ terms and the $y$ terms, and move the constant to the right.
$(x^{2} - 4 x) + (y^{2} + 6 y) = 3$
Complete each square and subtract the extra constants: $x^{2} - 4 x = {(x - 2)}^{2} - 4$ and $y^{2} + 6 y = {(y + 3)}^{2} - 9$ . Moving the $- 4$ and $- 9$ across adds them to the right-hand side.
${(x - 2)}^{2} + {(y + 3)}^{2} = 3 + 4 + 9 = 16$
Change the signs inside the brackets for the centre, and take the square root of $16$ for the radius: centre $(2, - 3)$ , radius $4$ . Check with the shortcut: $g = - 2$ , $f = 3$ , $c = - 3$ , so $r^{2} = 4 + 9 + 3 = 16$ .
For each point, find the squared distance from the centre and compare it with $r^{2} = 16$ . For $(5, 0)$ : ${(5 - 2)}^{2} + {(0 + 3)}^{2} = 9 + 9 = 18$ , and $18 > 16$ , so the point is outside.
For $(4, - 1)$ : ${(4 - 2)}^{2} + {(- 1 + 3)}^{2} = 4 + 4 = 8$ , and $8 < 16$ , so the point is inside.

Answer: Centre $(2, - 3)$ , radius $4$ . The point $(5, 0)$ is outside $C$ and the point $(4, - 1)$ is inside $C$ .

Example 2. The line $x + y = 5$ meets the circle $x^{2} + y^{2} - 2 x - 4 y - 5 = 0$ at the points $A$ and $B$ . Find the coordinates of $A$ and $B$ , and the exact length of $A B$ .

Show the working

Make $y$ the subject of the line: $y = 5 - x$ . Substitute into the circle with a bracket round every $y$ .
$x^{2} + {(5 - x)}^{2} - 2 x - 4 (5 - x) - 5 = 0$
Expand each term separately: ${(5 - x)}^{2} = 25 - 10 x + x^{2}$ and $- 4 (5 - x) = - 20 + 4 x$ .
$x^{2} + 25 - 10 x + x^{2} - 2 x - 20 + 4 x - 5 = 0$
Collect like terms. The constants cancel, leaving $2 x^{2} - 8 x = 0$ . Factorise; do not divide by $x$ , because that would lose the root $x = 0$ .
$2 x (x - 4) = 0$
So $x = 0$ or $x = 4$ . Two different roots, so the line meets the circle at two points. Find each $y$ from the line $y = 5 - x$ : $x = 0$ gives $y = 5$ , and $x = 4$ gives $y = 1$ .
Use the distance formula for $A B$ and simplify the surd.
$A B = \sqrt{{(4 - 0)}^{2} + {(1 - 5)}^{2}} = \sqrt{32} = 4 \sqrt{2}$
Check: the circle is ${(x - 1)}^{2} + {(y - 2)}^{2} = 10$ , with centre $(1, 2)$ . The midpoint of $A B$ is $(2, 3)$ and $A B$ has gradient $- 1$ , so its perpendicular bisector is $y = x + 1$ . That line passes through $(1, 2)$ , as the perpendicular bisector of a chord must.

Answer: $A (0, 5)$ and $B (4, 1)$ , in either order; $A B = 4 \sqrt{2}$ .

Example 3. The line $y = x + k$ is a tangent to the circle ${(x - 2)}^{2} + {(y + 1)}^{2} = 8$ . Find the two possible values of $k$ and the point of contact for each.

Show the working

There is no point of contact to start from, so use the intersection method. Substitute $y = x + k$ into the circle.
${(x - 2)}^{2} + {(x + k + 1)}^{2} = 8$
Expand, treating $k + 1$ as a single number: ${(x + k + 1)}^{2} = x^{2} + 2 (k + 1) x + {(k + 1)}^{2}$ . Collect into the form $a x^{2} + b x + c = 0$ , keeping $k$ inside the coefficients.
$2 x^{2} + (2 k - 2) x + (k^{2} + 2 k - 3) = 0$
A tangent gives a repeated root, so set $b^{2} - 4 a c = 0$ with $a = 2$ , $b = 2 k - 2$ and $c = k^{2} + 2 k - 3$ .
${(2 k - 2)}^{2} - 8 (k^{2} + 2 k - 3) = 0$
Expand: $4 k^{2} - 8 k + 4 - 8 k^{2} - 16 k + 24 = 0$ , so $- 4 k^{2} - 24 k + 28 = 0$ . Divide by $- 4$ and factorise.
$k^{2} + 6 k - 7 = 0 \Rightarrow (k + 7) (k - 1) = 0$
So $k = 1$ or $k = - 7$ : two parallel tangents with gradient $1$ , one on each side of the circle.
Put each $k$ back into the quadratic. $k = 1$ gives $2 x^{2} = 0$ , so $x = 0$ and $y = 0 + 1 = 1$ . $k = - 7$ gives $2 x^{2} - 16 x + 32 = 0$ , which is $2 {(x - 4)}^{2} = 0$ , so $x = 4$ and $y = 4 - 7 = - 3$ .
Check with the radius rule. From the centre $(2, - 1)$ to $(0, 1)$ the gradient is $\frac{2}{- 2} = - 1$ , and $- 1 \times 1 = - 1$ , so that radius is perpendicular to the tangent. From $(2, - 1)$ to $(4, - 3)$ the gradient is also $- 1$ .

Answer: $k = 1$ , touching at $(0, 1)$ ; $k = - 7$ , touching at $(4, - 3)$ .

Example 4. The circles $C_{1}$ : $x^{2} + y^{2} = 20$ and $C_{2}$ : $x^{2} + y^{2} - 10 x - 10 y + 40 = 0$ meet at the points $P$ and $Q$ . Find the equation of the common chord $P Q$ , and the coordinates of $P$ and $Q$ .

Show the working

Write $C_{1}$ as $x^{2} + y^{2} - 20 = 0$ , so that both equations have $0$ on the right and $x^{2}$ and $y^{2}$ with coefficient $1$ .
Subtract $C_{2}$ from $C_{1}$ . The $x^{2}$ and $y^{2}$ terms cancel.
$(x^{2} + y^{2} - 20) - (x^{2} + y^{2} - 10 x - 10 y + 40) = 0 \Rightarrow 10 x + 10 y - 60 = 0$
Divide by $10$ . The common chord is $x + y = 6$ .
Substitute $y = 6 - x$ into the simpler circle, $C_{1}$ .
$x^{2} + {(6 - x)}^{2} = 20 \Rightarrow 2 x^{2} - 12 x + 16 = 0$
Divide by $2$ : $x^{2} - 6 x + 8 = 0$ , so $(x - 2) (x - 4) = 0$ and $x = 2$ or $x = 4$ . Two roots confirm that the circles intersect at two points. Pair each with $y = 6 - x$ : $(2, 4)$ and $(4, 2)$ .
Check both points in $C_{2}$ : $4 + 16 - 20 - 40 + 40 = 0$ and $16 + 4 - 40 - 20 + 40 = 0$ . Also, the centres are $(0, 0)$ and $(5, 5)$ , so the line of centres has gradient $1$ , perpendicular to the chord's gradient $- 1$ .

Answer: The common chord is $x + y = 6$ ; $P$ and $Q$ are $(2, 4)$ and $(4, 2)$ .

Where people go wrong

Mistake: Reading the centre of ${(x + 3)}^{2} + {(y - 2)}^{2} = 16$ as $(3, - 2)$ .

Instead: The centre is $(- 3, 2)$ : change the sign of the number in each bracket.

The equation is ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ , so $x + 3$ means $x - (- 3)$ and $a = - 3$ .

Mistake: Giving the radius as the number on the right-hand side, e.g. $r = 16$ from ${(x - 1)}^{2} + y^{2} = 16$ .

Instead: The right-hand side is $r^{2}$ , so $r = 4$ . For $r^{2} = 20$ , $r = \sqrt{20} = 2 \sqrt{5}$ .

The equation comes from Pythagoras, where the hypotenuse, the radius, is squared.

Mistake: Completing the square without subtracting the extra constants: writing $(x^{2} - 6 x) + (y^{2} + 4 y) = 12$ as ${(x - 3)}^{2} + {(y + 2)}^{2} = 12$ .

Instead: ${(x - 3)}^{2} = x^{2} - 6 x + 9$ and ${(y + 2)}^{2} = y^{2} + 4 y + 4$ , so add $9$ and $4$ to the right-hand side: ${(x - 3)}^{2} + {(y + 2)}^{2} = 25$ .

The centre still comes out right, so the error is easy to miss, but the radius is wrong, and so is every later part that uses it.

Mistake: Using the gradient of the radius as the gradient of the tangent, or writing the tangent through the centre.

Instead: The tangent's gradient is the negative reciprocal of the radius's gradient, and the tangent passes through the point of contact $P$ .

The tangent is perpendicular to the radius at $P$ . It does not pass through the centre at all.

Mistake: Deciding that two circles intersect at two points because $d < r_{1} + r_{2}$ , without checking $| r_{1} - r_{2} |$ .

Instead: Two points of intersection need $| r_{1} - r_{2} | < d < r_{1} + r_{2}$ .

If $d < | r_{1} - r_{2} |$ , one circle lies inside the other and they never meet, even though $d < r_{1} + r_{2}$ is still true.

Check yourself

Write down the centre and the radius of the circle ${(x + 4)}^{2} + {(y - 1)}^{2} = 18$ .
Answer
Centre $(- 4, 1)$ ; radius $\sqrt{18} = 3 \sqrt{2}$ .
Find the centre and radius of the circle $x^{2} + y^{2} + 8 x - 10 y + 5 = 0$ .
Answer
${(x + 4)}^{2} - 16 + {(y - 5)}^{2} - 25 + 5 = 0$ , so ${(x + 4)}^{2} + {(y - 5)}^{2} = 36$ . Centre $(- 4, 5)$ , radius $6$ .
$A (- 1, 2)$ and $B (5, 10)$ are the ends of a diameter of a circle. Find the equation of the circle.
Answer
The centre is the midpoint, $(2, 6)$ . $A B = \sqrt{6^{2} + 8^{2}} = 10$ , so $r = 5$ . The equation is ${(x - 2)}^{2} + {(y - 6)}^{2} = 25$ .
Determine whether the line $y = x + 3$ is a tangent to the circle $x^{2} + y^{2} - 4 x - 6 y + 3 = 0$ , meets it at two points, or does not meet it. Give the coordinates of any points of intersection.
Answer
Substituting gives $2 x^{2} - 4 x - 6 = 0$ , so $x^{2} - 2 x - 3 = 0$ , with $b^{2} - 4 a c = 4 + 12 = 16 > 0$ . The line meets the circle at two points: $(x - 3) (x + 1) = 0$ gives $(3, 6)$ and $(- 1, 2)$ .
Show that the line $y = 3 - x$ does not meet the circle $x^{2} + y^{2} = 4$ .
Answer
$x^{2} + {(3 - x)}^{2} = 4$ gives $2 x^{2} - 6 x + 5 = 0$ . Then $b^{2} - 4 a c = 36 - 40 = - 4 < 0$ , so there are no real roots and the line does not meet the circle.
Find the equation of the tangent to the circle $x^{2} + y^{2} = 25$ at the point $(3, - 4)$ .
Answer
The radius has gradient $\frac{- 4 - 0}{3 - 0} = - \frac{4}{3}$ , so the tangent has gradient $\frac{3}{4}$ . Then $y + 4 = \frac{3}{4} (x - 3)$ , which gives $3 x - 4 y = 25$ .
Find the length of a tangent from the point $P (7, 5)$ to the circle ${(x - 1)}^{2} + {(y + 3)}^{2} = 36$ .
Answer
The centre is $C (1, - 3)$ , so $C P^{2} = 6^{2} + 8^{2} = 100$ . With $r^{2} = 36$ , the tangent length is $\sqrt{100 - 36} = 8$ .
Determine whether the circles $x^{2} + y^{2} = 4$ and ${(x - 3)}^{2} + {(y - 4)}^{2} = 16$ intersect, touch or do not intersect.
Answer
$d = \sqrt{3^{2} + 4^{2}} = 5$ , $r_{1} + r_{2} = 2 + 4 = 6$ and $| r_{1} - r_{2} | = 2$ . Since $2 < 5 < 6$ , the circles intersect at two points.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The circle $C$ has equation $x^{2} + y^{2} + 4 x - 2 y - 15 = 0$ .

(a) Find the coordinates of the centre of $C$ and the exact radius of $C$ . 3 marks No calculator

Mark scheme

M1 completes the square in both $x$ and $y$ , e.g. ${(x + 2)}^{2} - 4 + {(y - 1)}^{2} - 1 = 15$ , or uses centre $(- g, - f)$
A1 centre $(- 2, 1)$
A1 radius $\sqrt{20}$ or $2 \sqrt{5}$

(b) Verify that the point $P (0, 5)$ lies on $C$ . 1 mark No calculator

Mark scheme

B1 substitutes $(0, 5)$ to show $0 + 25 + 0 - 10 - 15 = 0$ , with a conclusion

(c) Find the equation of the tangent to $C$ at $P$ . Give your answer in the form $a x + b y = c$ , where $a$ , $b$ and $c$ are integers. 4 marks No calculator

Mark scheme

B1 gradient of the radius $C P$ is $\frac{5 - 1}{0 - (- 2)} = 2$ ; FT their centre
M1 tangent gradient $- \frac{1}{2}$ from their radius gradient, using $m_{1} m_{2} = - 1$
M1 equation of a straight line through $(0, 5)$ with their perpendicular gradient
A1 $x + 2 y = 10$ , or any integer multiple of it

Question 2. The line $y = m x$ , where $m$ is a constant, and the circle ${(x - 4)}^{2} + {(y - 2)}^{2} = 4$ are drawn on the same axes.

(a) Show that the $x$ -coordinates of any points where the line meets the circle satisfy $(1 + m^{2}) x^{2} - (8 + 4 m) x + 16 = 0$ . 2 marks No calculator

Mark scheme

M1 substitutes $y = m x$ into the circle: ${(x - 4)}^{2} + {(m x - 2)}^{2} = 4$
A1 expands both brackets correctly, $x^{2} - 8 x + 16 + m^{2} x^{2} - 4 m x + 4 = 4$ , and collects terms to reach the given equation; nfww

(b) Find the values of $m$ for which the line is a tangent to the circle. 3 marks No calculator

Mark scheme

M1 uses $b^{2} - 4 a c = 0$ with $a = 1 + m^{2}$ , $b = - (8 + 4 m)$ and $c = 16$
A1 $64 m - 48 m^{2} = 0$ oe, e.g. $16 m (4 - 3 m) = 0$
A1 $m = 0$ and $m = \frac{4}{3}$

Question 3. The circles $C_{1}$ and $C_{2}$ have equations $x^{2} + y^{2} = 10$ and $x^{2} + y^{2} - 8 x - 4 y + 10 = 0$ .

(a) Explain why $C_{1}$ and $C_{2}$ intersect at two points. 2 marks

Mark scheme

B1 both radii $\sqrt{10}$ , and the distance between the centres $(0, 0)$ and $(4, 2)$ is $\sqrt{20}$ oe
B1 comparison $0 < \sqrt{20} < 2 \sqrt{10}$ , e.g. via $20 < 40$ , so $| r_{1} - r_{2} | < d < r_{1} + r_{2}$ and the circles meet at two points

(b) Find the equation of the common chord of $C_{1}$ and $C_{2}$ . 2 marks

Mark scheme

M1 subtracts one equation from the other, both in expanded form, so that $x^{2}$ and $y^{2}$ cancel
A1 $2 x + y = 5$ oe

(c) Find the coordinates of the points where $C_{1}$ and $C_{2}$ intersect. 3 marks

Mark scheme

M1 substitutes their linear equation into either circle to obtain a quadratic in one variable
A1 $x^{2} - 4 x + 3 = 0$ oe, or $y^{2} - 2 y - 3 = 0$ oe if $x$ is eliminated
A1 $(1, 3)$ and $(3, - 1)$

How this is examined

Finding where a line meets a circle, or showing how it meets it: a Find the coordinates of the points where a line meets a circle given in either form; or a Determine whether a line is a tangent, cuts the circle or misses it The line rearranged and substituted with brackets (M1); a correct three-term quadratic equal to $0$ (A1); a valid solving method (M1); both points as coordinate pairs, each $y$ found from the line (A1, A1). For a Determine, the value of $b^{2} - 4 a c$ and a conclusion in words: ' $> 0$ , so two points', ' $= 0$ , so a tangent' or ' $< 0$ , so the line does not meet the circle'.
Deciding how two circles sit, then finding where they cross: an Explain or Determine about the distance between the centres, followed by a Find for the common chord or the points of intersection Both centres and radii, $d$ found exactly, and an explicit comparison with $r_{1} + r_{2}$ and $| r_{1} - r_{2} |$ that ends in one of the words intersect, touch or do not intersect. For the chord: the expanded equations subtracted to give a linear equation, then that line substituted into one circle and the quadratic solved, with each point written as a pair.
The equation of the tangent at a given point on the circle: a Find the equation of the tangent at $P$ , using a centre found in an earlier part, with the answer asked for in a stated form such as $a x + b y = c$ The gradient of the radius $C P$ (B1); the negative reciprocal for the tangent (M1); a straight line through $P$ , not the centre, with that gradient (M1); the final equation in exactly the form asked for, with integer coefficients if the question says so (A1).
Choosing a constant so that a line is a tangent: a Find the values of $k$ for which a line containing $k$ touches the circle, which may follow a Show that for the quadratic; or the inequality version, the values for which the line meets the circle at two points Substitution giving a quadratic in $x$ with $k$ in its coefficients (M1); $b^{2} - 4 a c = 0$ used with the correct $a$ , $b$ and $c$ (M1); a correct simplified equation in $k$ (A1); both values of $k$ (A1). For the inequality version, $b^{2} - 4 a c > 0$ and the correct region, between or outside the two critical values.
Centre and radius from an equation in expanded form: a Find or Write down at the start of a longer circle question Completing the square in $x$ and in $y$ with the extra constants subtracted, or using $(- g, - f)$ and $\sqrt{g^{2} + f^{2} - c}$ (M1); the centre with the correct signs (A1); the radius as $r$ , not $r^{2}$ , and as an exact surd when it is not a whole number (A1).

Words to know

Circle: The set of all points at a fixed distance, the radius, from a fixed point, the centre.
Chord: A straight line segment joining two points on a circle. A line that meets a circle at two points contains a chord.
Tangent: A line that touches a circle at exactly one point and is perpendicular to the radius there.
Point of contact: The single point where a tangent touches a circle, or where two touching circles meet.
Common chord: The chord joining the two points where two circles intersect. Its equation comes from subtracting the circles' expanded equations.
Discriminant: $b^{2} - 4 a c$ for $a x^{2} + b x + c = 0$ . After you substitute a line into a circle, its sign tells you whether the line cuts, touches or misses the circle.
Perpendicular bisector: The line at right angles to a segment through its midpoint. For any chord of a circle, it passes through the centre.

All Add Maths lessons Next: 9 Circular measure