What is 3.2: Circles?

A circle with centre (a,b) and radius r has equation (x-a)^2+(y-b)^2=r^2.

How many marks are 3.2 questions worth?

3.2 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has 3.2 appeared in?

We have not yet indexed a verified past-paper appearance for 3.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

3.2

Understand and use the coordinate geometry of the circle, including the equation (x − a)² + (y − b)² = r²; complete the square for centre and radius; use circle properties (semicircle angle, chord bisector, tangent-radius).

Draft — not yet indexed

Circles

Worked answers and methods for 3.2 on Edexcel A-level Maths 9MA0.

Explanation

A circle with centre $(a,b)$ and radius $r$ has equation $(x-a)^2+(y-b)^2=r^2$ .
Complete the square separately in $x$ and $y$ to reveal the centre and radius of a circle given in expanded form.
A radius is perpendicular to the tangent at its endpoint; perpendicular chord bisectors meet at the circumcentre, and an angle subtended by a diameter at the circumference is $90^\circ$ .
When testing tangency, require exactly one intersection or set the perpendicular distance from the centre to the line equal to the radius.
A common error is to read the centre as $(a,b)$ from $(x+a)^2+(y+b)^2$ instead of $(-a,-b)$ .

A point

(x,y)

lies on the circle when its distance from centre

(a,b)

r

Worked example

The point $P(5,3)$ lies on the circle with centre $(2,-1)$ . Determine the tangent at $P$ . Write your equation as $ax+by+c=0$ with integer coefficients.

1.The gradient of the radius from $(2,-1)$ to $(5,3)$ is $4/3$ , so the tangent gradient is $-3/4$ .
2.Hence $y-3=-\dfrac{3}{4}(x-5)$ .
3.Multiplying by $4$ and rearranging gives $3x+4y-27=0$ .

Answer: $3x+4y-27=0$

Common mistakes

Don't complete the square but treat $r^2$ as the radius instead of taking its positive square root.
Don't find the radius gradient but forget that the tangent gradient is its negative reciprocal.

Exam tip

For a tangent equation, join the centre to the contact point first, then use perpendicular gradients.

Worked practice

Tier 1 · Easy

Find the centre and radius of the circle

x^2+y^2-6x+8y-11=0

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Centre $(3,-4)$ Radius $6$	3
Notes
Complete both squares: $x^2-6x=(x-3)^2-9$ and $y^2+8y=(y+4)^2-16$ . The equation becomes $(x-3)^2+(y+4)^2=36$ , so the centre is $(3,-4)$ and the radius is $6$ .

(3 marks)

Tier 2 · Standard

The points

A(-2,1)

and

B(6,5)

are the endpoints of a diameter of a circle. Find the equation of the circle.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$(x-2)^2+(y-3)^2=20$	4
Notes
The centre is the midpoint of $AB$ , namely $((-2+6)/2,(1+5)/2)=(2,3)$ . The squared radius is the squared distance from the centre to $A$ : $(-2-2)^2+(1-3)^2=16+4=20$ . Hence the circle is $(x-2)^2+(y-3)^2=20$ .

(4 marks)

Tier 3 · Hard

The points

A(-1,2)

B(5,2)

and

C(1,6)

lie on a circle. Use perpendicular bisectors to find the equation of this circumcircle.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$(x-2)^2+(y-3)^2=10$	5
Notes
The midpoint of horizontal chord $AB$ is $(2,2)$ , so its perpendicular bisector is $x=2$ . The midpoint of $AC$ is $(0,4)$ and the gradient of $AC$ is $2$ , so its perpendicular bisector is $y-4=-\dfrac{1}{2}x$ . At $x=2$ this gives $y=3$ , so the centre is $(2,3)$ . The squared radius is $(-1-2)^2+(2-3)^2=10$ , hence the circumcircle is $(x-2)^2+(y-3)^2=10$ .

(5 marks)

Tier 1 · Easy

A circle has centre

(-3,4)

and passes through

(1,1)

. Find its equation.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$(x+3)^2+(y-4)^2=25$	2
Notes
The squared radius is $(1+3)^2+(1-4)^2=16+9=25$ . Therefore the circle is $(x+3)^2+(y-4)^2=25$ .

(2 marks)

Tier 2 · Standard

For the circle

C

with equation

x^2+y^2+4x-6y-12=0

, determine both

x

-axis intercepts. At the intercept whose

x

-coordinate is positive, find the tangent and write it as

ax+by+c=0

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$(-6,0)$ and $(2,0)$ $4x-3y-8=0$	5
Notes
On the $x$ -axis, $y=0$ , so $x^2+4x-12=0=(x+6)(x-2)$ , giving $(-6,0)$ and $(2,0)$ . Completing the square gives $(x+2)^2+(y-3)^2=25$ , so the centre is $(-2,3)$ . The radius to $(2,0)$ has gradient $-3/4$ , hence the tangent gradient is $4/3$ . Thus $y=\dfrac43(x-2)$ , or $4x-3y-8=0$ .

(5 marks)

Tier 3 · Hard

The circle

C

has centre

(2,-1)

and radius

5

. The point

P(15,-1)

lies outside

C

. Find the equations of both tangents from

P

C

and the exact coordinates of both points of contact.

(7)

(Total for Question 6 is 7 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$5x-12y-87=0$ , touching at $\left(\dfrac{51}{13},-\dfrac{73}{13}\right)$ $5x+12y-63=0$ , touching at $\left(\dfrac{51}{13},\dfrac{47}{13}\right)$	7
Notes
Let a tangent through $P$ have equation $y+1=m(x-15)$ . Its perpendicular distance from the centre $(2,-1)$ must be $5$ , so $\|13m\|/\sqrt{m^2+1}=5$ . Hence $169m^2=25m^2+25$ , giving $m=\pm5/12$ . The two tangent equations are therefore $5x-12y-87=0$ and $5x+12y-63=0$ . The point of contact is the foot of the perpendicular from the centre. Solving each tangent together with its perpendicular through $(2,-1)$ gives respectively $\left(51/13,-73/13\right)$ and $\left(51/13,47/13\right)$ .

(7 marks)

Tier 2 · Standard

The points

A(-4,0)

and

B(6,0)

are the endpoints of a diameter of a circle. The point

C=(p,4)

lies on the circle. Using the angle in a semicircle, find all possible values of

p

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$p=-2$ or $p=4$	4
Notes
The angle $ACB$ is $90^\circ$ , so $AC$ and $BC$ are perpendicular. The cases $p=-4$ and $p=6$ give one vertical side but the other side is not horizontal, so neither works. Otherwise the gradients are $4/(p+4)$ and $-4/(6-p)$ . Hence $-16/[(p+4)(6-p)]=-1$ , so $(p+4)(6-p)=16$ . This simplifies to $p^2-2p-8=0=(p+2)(p-4)$ , giving $p=-2$ or $p=4$ .

(4 marks)

Tier 3 · Hard

The circles

C_1

and

C_2

have equations

x^2+y^2=25

and

(x-6)^2+y^2=25

respectively. Find an equation of their common chord and the exact coordinates of its endpoints. Hence find the equation of the circle whose diameter is this common chord.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$x=3$ $(3,4)$ and $(3,-4)$ $(x-3)^2+y^2=16$	6
Notes
Subtracting the two circle equations gives $x^2-(x-6)^2=0$ , hence $12x-36=0$ and the common chord is $x=3$ . Substitution into $x^2+y^2=25$ gives $9+y^2=25$ , so its endpoints are $(3,4)$ and $(3,-4)$ . Their midpoint is $(3,0)$ and half their separation is $4$ , so the circle with this chord as diameter is $(x-3)^2+y^2=16$ .

(6 marks)

Tier 3 · Hard

A circle has its centre in the first quadrant and is tangent to both coordinate axes. It passes through the point

(6,8)

. Find the equations of all possible circles, giving the radii in exact form.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$(x-r)^2+(y-r)^2=r^2$ , where $r=14-4\sqrt6$ or $r=14+4\sqrt6$	5
Notes
Tangency to both positive coordinate axes means that the centre is $(r,r)$ and the radius is $r>0$ . Since $(6,8)$ lies on the circle, $(6-r)^2+(8-r)^2=r^2$ . This simplifies to $r^2-28r+100=0$ , so $r=[28\pm\sqrt{384}]/2=14\pm4\sqrt6$ . Both values are positive, so both give valid circles. Substituting either exact value for $r$ in $(x-r)^2+(y-r)^2=r^2$ gives the two equations.

(5 marks)

Q10

Tier 3 · Hard

10.

The line

y=2x+c

cuts the circle

(x+1)^2+(y-3)^2=49

in a chord of length

6

. Find all possible values of

c

. For each value, find the exact coordinates of the midpoint of the chord.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$c=5+10\sqrt2$ , midpoint $\left(-1-4\sqrt2,3+2\sqrt2\right)$ $c=5-10\sqrt2$ , midpoint $\left(-1+4\sqrt2,3-2\sqrt2\right)$	6
Notes
The circle has centre $(-1,3)$ and radius $7$ . If $d$ is the perpendicular distance from the centre to the chord, then $d^2+3^2=7^2$ , so $d=2\sqrt{10}$ . Writing the line as $2x-y+c=0$ , its distance from the centre is $\|c-5\|/\sqrt5$ . Thus $\|c-5\|/\sqrt5=2\sqrt{10}$ , giving $c=5\pm10\sqrt2$ . The chord midpoint is the foot of the perpendicular from $(-1,3)$ to the line. If $q=c-5$ , this foot is $(-1-2q/5,3+q/5)$ , which gives the two stated midpoints.

(6 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for 3.2. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in 3 Coordinate geometry in the (x, y) plane

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