Circles – A Level Maths Year 1

1 The Equation of a Circle

A circle is the set of all points at a fixed distance $r$ (the radius) from a fixed point (the centre). Writing that sentence with the distance formula gives the equation. This page completes DfE content area C · Coordinate geometry.

Circle with centre $(a, b)$ and radius $r$

$(x - a)^2 + (y - b)^2 = r^2$

Where it comes from. A point $(x, y)$ is on the circle exactly when its distance from $(a, b)$ is $r$: $\sqrt{(x-a)^2 + (y-b)^2} = r$. Squaring both sides gives the formula. With centre the origin it becomes $x^2 + y^2 = r^2$.

Signs inside the brackets are opposite to the centre. $(x + 4)^2 + (y - 1)^2 = 9$ has centre $(-4, 1)$, and radius $3$ — not $9$.

Worked Example 1 — Writing down the equation

Find the equation of the circle with centre $(3, -2)$ and radius $5$, and show that $(6, 2)$ lies on it.

①$(x - 3)^2 + (y + 2)^2 = 25$

②At $(6, 2)$: $(6 - 3)^2 + (2 + 2)^2 = 9 + 16 = 25$ ✓

Inside, on or outside? Substitute the point into $(x-a)^2 + (y-b)^2$ and compare with $r^2$: less means inside, equal means on, greater means outside.

2 Finding the Centre and Radius by Completing the Square

If you expand $(x-3)^2 + (y+2)^2 = 25$ you get $x^2 + y^2 - 6x + 4y - 12 = 0$, which hides the centre and radius. To get them back, complete the square in $x$ and in $y$ separately.

Group the $x$ terms and the $y$ terms. (If $x^2$ and $y^2$ have a common coefficient other than $1$, divide through first.)
Complete the square on each: $x^2 + px \equiv \left(x + \tfrac{p}{2}\right)^2 - \left(\tfrac{p}{2}\right)^2$.
Move all the constants to the right-hand side. That number is $r^2$.

Worked Example 2 — Centre and radius

Find the centre and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$.

①$x^2 - 6x = (x - 3)^2 - 9$ and $y^2 + 4y = (y + 2)^2 - 4$

②$(x-3)^2 - 9 + (y+2)^2 - 4 - 12 = 0$

③$(x-3)^2 + (y+2)^2 = 25$

④Centre $(3, -2)$, radius $5$.

Worked Example 3 — Divide first, and when is it a circle?

(a) Find the centre and radius of $2x^2 + 2y^2 + 8x - 12y + 8 = 0$. (b) For what values of $k$ does $x^2 + y^2 + 2x - 4y + k = 0$ represent a circle?

①(a) Divide by $2$: $x^2 + y^2 + 4x - 6y + 4 = 0$

②$(x+2)^2 - 4 + (y-3)^2 - 9 + 4 = 0 \;\Rightarrow\; (x+2)^2 + (y-3)^2 = 9$

③Centre $(-2, 3)$, radius $3$.

④(b) $(x+1)^2 - 1 + (y-2)^2 - 4 + k = 0 \;\Rightarrow\; (x+1)^2 + (y-2)^2 = 5 - k$

⑤A circle needs $r^2 > 0$, so $5 - k > 0$, i.e. $k < 5$.

Only if the coefficients of $x^2$ and $y^2$ are equal (and there is no $xy$ term) can the equation be a circle. $x^2 + 2y^2 = 9$ is not a circle.

3 Three Circle Theorems in Coordinates

The subject content names three properties from GCSE geometry that you must be able to use with coordinates.

Property	How it is used
The angle in a semicircle is a right angle	If $AB$ is a diameter and $C$ is on the circle, then $AC \perp BC$. Conversely, if $\angle ACB = 90°$, then $AB$ is a diameter.
The perpendicular from the centre to a chord bisects the chord	The perpendicular bisector of any chord passes through the centre.
The radius is perpendicular to the tangent at the point of contact	Tangent gradient $= -\dfrac{1}{\text{radius gradient}}$.

Worked Example 4 — The tangent at a point

Find the equation of the tangent to $(x-3)^2 + (y+2)^2 = 25$ at $P(6, 2)$.

①Gradient of the radius $CP$: $\dfrac{2 - (-2)}{6 - 3} = \dfrac43$

②The tangent is perpendicular to the radius: gradient $-\tfrac34$.

③$y - 2 = -\tfrac34(x - 6) \;\Rightarrow\; 4y - 8 = -3x + 18$

④$3x + 4y - 26 = 0$ (the red line in the diagram above).

Worked Example 5 — Angle in a semicircle

$A(-1, 2)$ and $B(5, 10)$ are the ends of a diameter. Find the equation of the circle, and show that $\angle ACB = 90°$ where $C(6, 9)$.

①Centre = midpoint of $AB$ = $(2, 6)$.

②$r^2 = (5 - 2)^2 + (10 - 6)^2 = 9 + 16 = 25$, so $(x-2)^2 + (y-6)^2 = 25$.

③$C$: $(6-2)^2 + (9-6)^2 = 16 + 9 = 25$ ✓, so $C$ is on the circle.

④$m_{AC} = \dfrac{9 - 2}{6 + 1} = 1$, $m_{BC} = \dfrac{9 - 10}{6 - 5} = -1$. Product $-1$, so $\angle ACB = 90°$ ✓

Worked Example 6 — The circle through three points

Find the equation of the circle through $A(0, 5)$, $B(7, 4)$ and $C(8, 1)$.

①Chord $AB$: midpoint $(3.5, 4.5)$, gradient $-\tfrac17$, so perpendicular gradient $7$. Bisector: $y - 4.5 = 7(x - 3.5)$, i.e. $y = 7x - 20$.

②Chord $BC$: midpoint $(7.5, 2.5)$, gradient $\dfrac{1 - 4}{8 - 7} = -3$, so perpendicular gradient $\tfrac13$. Bisector: $y - 2.5 = \tfrac13(x - 7.5)$, i.e. $y = \tfrac13 x$.

③Both bisectors pass through the centre: $7x - 20 = \tfrac13 x \;\Rightarrow\; 21x - 60 = x \;\Rightarrow\; x = 3$, $y = 1$.

④$r^2 = (0 - 3)^2 + (5 - 1)^2 = 25$, so $(x-3)^2 + (y-1)^2 = 25$.

⑤Check $C$: $25 + 0 = 25$ ✓

4 Where a Line Meets a Circle

Substitute the line into the circle. You get a quadratic, and its discriminant tells you how many points of intersection there are.

Discriminant $b^2 - 4ac$	Number of intersections	Geometry
$> 0$	Two	The line is a chord (a secant)
$= 0$	One (repeated root)	The line is a tangent
$< 0$	None	The line misses the circle

Worked Example 7 — Points of intersection

Find where the line $y = x + 1$ meets the circle $x^2 + y^2 = 25$.

①$x^2 + (x+1)^2 = 25 \;\Rightarrow\; 2x^2 + 2x - 24 = 0 \;\Rightarrow\; x^2 + x - 12 = 0$

②$(x + 4)(x - 3) = 0$, so $x = 3$ or $x = -4$.

③Points: $(3, 4)$ and $(-4, -3)$.

Worked Example 8 — A tangency condition

Find the values of $k$ for which $y = x + k$ is a tangent to $x^2 + y^2 = 8$.

①$x^2 + (x + k)^2 = 8 \;\Rightarrow\; 2x^2 + 2kx + k^2 - 8 = 0$

②Tangent $\iff$ discriminant $= 0$: $(2k)^2 - 4(2)(k^2 - 8) = 0$

③$4k^2 - 8k^2 + 64 = 0 \;\Rightarrow\; k^2 = 16$

④$k = 4$ or $k = -4$ — one tangent on each side of the circle.

Worked Example 9 — A chord and its midpoint

The line $y = 1$ cuts the circle $(x-3)^2 + (y+2)^2 = 25$ at $P$ and $Q$. Find the length $PQ$ and verify that the perpendicular from the centre bisects $PQ$.

①$(x - 3)^2 + 9 = 25 \;\Rightarrow\; (x-3)^2 = 16 \;\Rightarrow\; x = 7$ or $x = -1$.

②$P(-1, 1)$, $Q(7, 1)$, so $PQ = 8$.

③The perpendicular from $C(3, -2)$ to the horizontal chord is the vertical line $x = 3$, meeting it at $(3, 1)$ — the midpoint of $PQ$ ✓

"Show detailed reasoning" questions on intersections expect the substitution, the quadratic and the discriminant written out. A calculator's equation solver is fine for checking, but not as the method.

5 Quick Reference

Standard form

$(x-a)^2 + (y-b)^2 = r^2$

Centre signs

Opposite to the signs in the brackets.

Expanded form

Complete the square in $x$ and $y$; the right side is $r^2$.

Is it a circle?

Need $r^2 > 0$.

Semicircle

Diameter $AB$, $C$ on circle $\Rightarrow AC \perp BC$.

Chord

Its perpendicular bisector passes through the centre.

Tangent

Perpendicular to the radius at the contact point.

Line meets circle

Substitute; the discriminant counts the intersections.

Tangent condition

$b^2 - 4ac = 0$.

Point test

Compare $(x-a)^2 + (y-b)^2$ with $r^2$.

6 Practice Questions

Question 1

Write down the centre and radius of $(x + 4)^2 + (y - 1)^2 = 49$.

▶ Show solution

Centre $(-4, 1)$, radius $\sqrt{49} = 7$.

Question 2

Find the equation of the circle with centre $(-1, 5)$ that passes through $(2, 1)$.

▶ Show solution

$r^2 = (2 + 1)^2 + (1 - 5)^2 = 9 + 16 = 25$

$(x + 1)^2 + (y - 5)^2 = 25$

Question 3

Find the centre and radius of $x^2 + y^2 + 10x - 8y + 5 = 0$.

▶ Show solution

$(x + 5)^2 - 25 + (y - 4)^2 - 16 + 5 = 0$

$(x+5)^2 + (y-4)^2 = 36$

Centre $(-5, 4)$, radius $6$.

Question 4

$(-3, 4)$ and $(5, -2)$ are the ends of a diameter of a circle. Find its equation.

▶ Show solution

Centre = midpoint $= (1, 1)$.

$r^2 = (5 - 1)^2 + (-2 - 1)^2 = 16 + 9 = 25$

$(x - 1)^2 + (y - 1)^2 = 25$

Question 5

Show that $(5, 7)$ lies on the circle $x^2 + y^2 - 4x - 6y - 12 = 0$ and find the equation of the tangent there in the form $ax + by + c = 0$.

▶ Show solution

$25 + 49 - 20 - 42 - 12 = 0$ ✓, so $(5, 7)$ is on the circle.

Completing the square: $(x - 2)^2 + (y - 3)^2 = 25$, centre $(2, 3)$.

Radius gradient $\dfrac{7 - 3}{5 - 2} = \dfrac43$, so tangent gradient $-\tfrac34$.

$y - 7 = -\tfrac34(x - 5) \;\Rightarrow\; 4y - 28 = -3x + 15$

$3x + 4y - 43 = 0$

Question 6

Find the coordinates of the points where $y = x + 2$ meets $x^2 + y^2 = 10$.

▶ Show solution

$x^2 + (x + 2)^2 = 10 \;\Rightarrow\; 2x^2 + 4x - 6 = 0 \;\Rightarrow\; x^2 + 2x - 3 = 0$

$(x + 3)(x - 1) = 0$, so $x = 1$ or $x = -3$.

Points $(1, 3)$ and $(-3, -1)$.

Question 7

In this question you must show detailed reasoning. Show that the line $x + y = 10$ does not meet the circle $(x - 2)^2 + (y - 1)^2 = 4$.

▶ Show solution

Substitute $y = 10 - x$: $(x - 2)^2 + (9 - x)^2 = 4$

$x^2 - 4x + 4 + x^2 - 18x + 81 - 4 = 0 \;\Rightarrow\; 2x^2 - 22x + 81 = 0$

Discriminant: $(-22)^2 - 4(2)(81) = 484 - 648 = -164 < 0$

No real roots, so the line and circle do not meet.

Question 8

The line $y = mx$ is a tangent to the circle $(x - 4)^2 + y^2 = 4$. Find the exact possible values of $m$.

▶ Show solution

$(x - 4)^2 + m^2x^2 = 4 \;\Rightarrow\; (1 + m^2)x^2 - 8x + 12 = 0$

Tangent: $64 - 4(1 + m^2)(12) = 0 \;\Rightarrow\; 1 + m^2 = \tfrac{64}{48} = \tfrac43$

$m^2 = \tfrac13$, so $m = \pm\dfrac{1}{\sqrt3} = \pm\dfrac{\sqrt3}{3}$.

Question 9

The equation $x^2 + y^2 - 6x + 2y + k = 0$ represents a circle. (a) Find the range of possible values of $k$. (b) Find $k$ if the circle touches the $x$-axis.

▶ Show solution

(a) $(x - 3)^2 - 9 + (y + 1)^2 - 1 + k = 0 \;\Rightarrow\; (x-3)^2 + (y+1)^2 = 10 - k$.

Need $10 - k > 0$, so $k < 10$.

(b) The centre $(3, -1)$ is $1$ unit from the $x$-axis, so touching the axis means $r = 1$.

$10 - k = 1$, so $k = 9$.

Question 10

$A(-2, -2)$, $B(6, 4)$ and $C(5, 5)$.

(a) Show that $\angle ACB = 90°$. (b) Hence find the equation of the circle through $A$, $B$ and $C$. (c) Find the equation of the tangent to the circle at $C$. (d) The tangent meets the $x$-axis at $T$. Find the exact length $CT$.

▶ Show solution

(a) $m_{AC} = \dfrac{5 + 2}{5 + 2} = 1$, $m_{BC} = \dfrac{5 - 4}{5 - 6} = -1$. Product $-1$, so $\angle ACB = 90°$.

(b) The angle in a semicircle is $90°$, so $AB$ is a diameter. Centre $M$ = midpoint of $AB$ = $(2, 1)$.

$r = \tfrac12 AB = \tfrac12\sqrt{8^2 + 6^2} = \tfrac12(10) = 5$.

$(x - 2)^2 + (y - 1)^2 = 25$. (Check $C$: $9 + 16 = 25$ ✓)

(c) $m_{MC} = \dfrac{5 - 1}{5 - 2} = \dfrac43$, so the tangent gradient is $-\tfrac34$.

$y - 5 = -\tfrac34(x - 5) \;\Rightarrow\; 3x + 4y - 35 = 0$

(d) $y = 0$: $3x = 35$, so $T\left(\tfrac{35}{3}, 0\right)$.

$CT^2 = \left(\tfrac{35}{3} - 5\right)^2 + (0 - 5)^2 = \tfrac{400}{9} + \tfrac{225}{9} = \tfrac{625}{9}$, so $CT = \tfrac{25}{3}$.

Check: triangle $MCT$ has a right angle at $C$, so $MT^2 = 5^2 + \left(\tfrac{25}{3}\right)^2 = \tfrac{850}{9}$, and directly $MT^2 = \left(\tfrac{29}{3}\right)^2 + 1^2 = \tfrac{850}{9}$ ✓

Circles · A Level Maths Year 1 · Created with MathJax