The Geometry of Circles – OCR FSMQ Additional Maths

1 The Equation of a Circle

CG3 — as the specification writes it

$(x-a)^2 + (y-b)^2 = r^2$
where $(a, b)$ is the centre and $r$ is the radius

Why it looks like that. A circle is the set of points at distance $r$ from the centre. Apply the distance formula from $(x,y)$ to $(a,b)$, set it equal to $r$, and square both sides — the circle equation is just Pythagoras in disguise.

The signs are opposite. $(x-3)^2 + (y+5)^2 = 49$ has centre $(3, -5)$ and radius $7$. Rewrite $(y+5)$ as $(y - (-5))$ if that helps you see it.

Equation	Centre	Radius
$x^2 + y^2 = 25$	$(0, 0)$	$5$
$(x-2)^2 + (y-7)^2 = 9$	$(2, 7)$	$3$
$(x+1)^2 + (y-4)^2 = 36$	$(-1, 4)$	$6$
$(x+3)^2 + (y+2)^2 = 20$	$(-3, -2)$	$2\sqrt5$
$x^2 + (y-6)^2 = 1$	$(0, 6)$	$1$

2 Finding a Circle's Equation

Worked Example 1 — Centre and a point

A circle has centre $C(3, -1)$ and passes through $P(7, 2)$. Find its equation.

①The radius is the distance $CP$: $r = \sqrt{(7-3)^2 + (2-(-1))^2}$

②$= \sqrt{16 + 9} = \sqrt{25} = 5$

③$(x-3)^2 + (y+1)^2 = 25$

④Check with $P$: $(7-3)^2 + (2+1)^2 = 16 + 9 = 25$ ✓

You never need to find $r$ itself — only $r^2$. If the radius works out as $\sqrt{13}$, just write $r^2 = 13$ and move on.

Worked Example 2 — The specification's own example

$A(1, 1)$ and $B(5, 7)$ are the ends of a diameter. Show that the equation of the circle is $(x-3)^2 + (y-4)^2 = 13$.

①The centre is the mid-point of the diameter: $\left(\tfrac{1+5}{2}, \tfrac{1+7}{2}\right) = (3, 4)$

②The diameter length is $AB = \sqrt{4^2 + 6^2} = \sqrt{52}$

③So $r = \tfrac12\sqrt{52}$, and $r^2 = \tfrac14 \times 52 = 13$.

④$(x-3)^2 + (y-4)^2 = 13$ as required.

⑤Faster route to $r^2$: use the distance from the centre to $A$ directly: $(1-3)^2 + (1-4)^2 = 4 + 9 = 13$ ✓

Step ⑤ is the method to use. Going via the full diameter and halving invites arithmetic errors with the surds. Find the centre, then use one endpoint.

3 Recognising the Expanded Form

A circle equation is often given multiplied out. To find the centre and radius you must complete the square in $x$ and in $y$ separately.

Collect the $x$ terms together and the $y$ terms together.
Complete the square on the $x$ terms, and on the $y$ terms.
Move the two leftover constants to the right-hand side.
Read off the centre and $r^2$.

Worked Example 3 — From expanded to standard form

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

①Group: $\left(x^2 - 6x\right) + \left(y^2 + 4y\right) = 12$

②$x^2 - 6x = (x-3)^2 - 9$

③$y^2 + 4y = (y+2)^2 - 4$

④$(x-3)^2 - 9 + (y+2)^2 - 4 = 12$

⑤$(x-3)^2 + (y+2)^2 = 25$

⑥Centre $(3, -2)$, radius $5$.

Both leftover constants move across. Forgetting the $-4$ from the $y$ terms is the standard slip, and it changes the radius.

A quick recognition test. An equation is a circle if the $x^2$ and $y^2$ coefficients are equal and there is no $xy$ term. If the coefficients differ, it is an ellipse — beyond this course.

4 Points, Lines and Circles

Is a point inside, on or outside?

Substitute into the left-hand side and compare with $r^2$:
less than $r^2$ → inside · equal → on · greater → outside

Worked Example 4 — Where a line meets a circle

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

①Substitute: $x^2 + (x+1)^2 = 25$

②$x^2 + x^2 + 2x + 1 = 25$

③$2x^2 + 2x - 24 = 0$, so $x^2 + x - 12 = 0$

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

⑤$x = -4 \Rightarrow y = -3$; $x = 3 \Rightarrow y = 4$

⑥The line cuts the circle at $(-4, -3)$ and $(3, 4)$.

The discriminant decides the geometry

Two solutions → the line cuts the circle
One (repeated) solution → the line is a tangent
No real solutions → the line misses the circle

Worked Example 5 — Proving a tangent

Show that $y = 2x - 10$ is a tangent to $x^2 + y^2 = 20$, and find the point of contact.

①$x^2 + (2x-10)^2 = 20$

②$x^2 + 4x^2 - 40x + 100 = 20$

③$5x^2 - 40x + 80 = 0$, so $x^2 - 8x + 16 = 0$

④$\Delta = 64 - 64 = 0$ — exactly one solution, so the line is a tangent.

⑤$(x-4)^2 = 0$ gives $x = 4$, and $y = 2(4) - 10 = -2$.

⑥Point of contact $(4, -2)$.

⑦Check: $16 + 4 = 20$ ✓ and the radius to $(4,-2)$ has gradient $-\tfrac12$, whose negative reciprocal is $2$ — the line's gradient ✓

The tangent–radius property

A tangent is always perpendicular to the radius at the point of contact

Worked Example 6 — A tangent at a given point

Find the equation of the tangent to $(x-1)^2 + (y-2)^2 = 25$ at the point $P(5, 5)$.

①Confirm $P$ is on the circle: $(5-1)^2 + (5-2)^2 = 16 + 9 = 25$ ✓

②Centre $C(1, 2)$. Gradient of $CP = \dfrac{5-2}{5-1} = \dfrac34$

③The tangent is perpendicular, so its gradient is $-\dfrac43$.

④$y - 5 = -\tfrac43(x - 5)$

⑤$3y - 15 = -4x + 20$, so $4x + 3y = 35$.

There is no need for calculus here. The tangent–radius property is faster and always available for circles.

5 Quick Reference

The equation

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

Centre

Opposite signs to the brackets.

Radius

Square-root the right-hand side.

Diameter given

Centre is the mid-point; use one endpoint for $r^2$.

Expanded form

Complete the square in $x$ and in $y$.

Recognition

Equal $x^2$, $y^2$ coefficients and no $xy$ term.

Point test

Substitute and compare with $r^2$.

Line and circle

Substitute, then use the discriminant.

$\Delta = 0$

Tangent.

Tangent gradient

Negative reciprocal of the radius gradient.

6 Practice Questions

Question 1

Write down the centre and radius of $(x-4)^2 + (y+3)^2 = 49$.

▶ Show solution

Centre $(4, -3)$, radius $7$.

Question 2

Find the equation of the circle with centre $(-2, 5)$ and radius $4$.

▶ Show solution

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

Question 3

A circle has centre $(1, 3)$ and passes through $(5, 6)$. Find its equation.

▶ Show solution

$r^2 = (5-1)^2 + (6-3)^2 = 16 + 9 = 25$

$(x-1)^2 + (y-3)^2 = 25$

Question 4

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

▶ Show solution

$(x^2 - 8x) + (y^2 + 2y) = -8$

$(x-4)^2 - 16 + (y+1)^2 - 1 = -8$

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

Centre $(4, -1)$, radius $3$.

Question 5

$A(2, 1)$ and $B(8, 9)$ are ends of a diameter. Find the circle's equation.

▶ Show solution

Centre: mid-point $= (5, 5)$.

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

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

Question 6

Is $(1, 6)$ inside, on or outside the circle $(x-3)^2 + (y-2)^2 = 20$?

▶ Show solution

$(1-3)^2 + (6-2)^2 = 4 + 16 = 20$

This equals $r^2$, so the point lies on the circle.

Question 7

Find where the line $y = x - 1$ meets the circle $x^2 + y^2 = 13$.

▶ Show solution

$x^2 + (x-1)^2 = 13$

$2x^2 - 2x + 1 = 13$, so $2x^2 - 2x - 12 = 0$, i.e. $x^2 - x - 6 = 0$.

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

Points: $(3, 2)$ and $(-2, -3)$.

Question 8

Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(3, 4)$.

▶ Show solution

Centre is the origin. Gradient of the radius $= \dfrac{4}{3}$.

Tangent gradient $= -\dfrac34$.

$y - 4 = -\tfrac34(x-3)$

$4y - 16 = -3x + 9$, so $3x + 4y = 25$.

Question 9

Show that the line $y = x + 6$ does not meet the circle $x^2 + y^2 = 9$.

▶ Show solution

$x^2 + (x+6)^2 = 9$

$2x^2 + 12x + 36 = 9$, so $2x^2 + 12x + 27 = 0$.

$\Delta = 144 - 4(2)(27) = 144 - 216 = -72$

The discriminant is negative, so there are no real solutions — the line misses the circle.

Geometrically: the perpendicular distance from the origin to the line $x - y + 6 = 0$ is $\dfrac{6}{\sqrt2} = 3\sqrt2 \approx 4.24$, which is greater than the radius $3$ ✓

Question 10

A circle passes through $A(0, 0)$, $B(6, 0)$ and $C(0, 8)$.

(a) Explain why $BC$ must be a diameter. (b) Find the centre and radius. (c) Find the circle's equation in expanded form. (d) Find the equation of the tangent at $A$.

▶ Show solution

(a) $AB$ lies along the $x$-axis and $AC$ along the $y$-axis, so the angle $\angle BAC$ is $90°$.

An angle of $90°$ in a circle stands on a diameter (the angle in a semicircle), so $BC$ is a diameter.

(b) The centre is the mid-point of $BC$: $\left(\tfrac{6+0}{2}, \tfrac{0+8}{2}\right) = (3, 4)$.

$r^2 = (0-3)^2 + (0-4)^2 = 9 + 16 = 25$, so $r = 5$.

Check the other points: $B$: $(6-3)^2 + (0-4)^2 = 9+16 = 25$ ✓ $C$: $(0-3)^2+(8-4)^2 = 25$ ✓

(c) $(x-3)^2 + (y-4)^2 = 25$

$x^2 - 6x + 9 + y^2 - 8y + 16 = 25$

$x^2 + y^2 - 6x - 8y = 0$

The absence of a constant term is the signature of a circle through the origin — as a check, substituting $(0,0)$ gives $0 = 0$ ✓

(d) The radius from the centre $(3,4)$ to $A(0,0)$ has gradient $\dfrac{4}{3}$.

The tangent at $A$ is perpendicular, gradient $-\dfrac34$, and passes through the origin:

$y = -\dfrac{3}{4}x$, or $3x + 4y = 0$.

The Geometry of Circles (CG3) · OCR FSMQ Additional Maths · Created with MathJax