What is A16: Equation of a circle?

Higher tier only. A circle centred at (0,0) with radius r has equation x^2+y^2=r^2.

How many marks are A16 questions worth?

In verified past-paper appearances, A16 questions on Edexcel GCSE Maths 1MA1 have been worth 3–5 marks.

Which past exam papers has A16 appeared in?

A16 has appeared in 7 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

A16

Recognise and use the equation of a circle with centre at the origin; find the equation of a tangent to a circle at a given point [Higher only]

Higher only

Equation of a circle

Worked answers, methods and verified real exam appearances for A16 on Edexcel GCSE Maths 1MA1.

Explanation

Higher tier only. A circle centred at $(0,0)$ with radius $r$ has equation $x^2+y^2=r^2$ .
Check that a given point lies on the circle by substituting its coordinates. The radius from the origin to a point $(a,b)$ has gradient $b/a$ when $a\ne0$ .
The tangent at that point is perpendicular to the radius, so its gradient is the negative reciprocal, and a point-gradient equation gives the tangent.
The examiner expects an exact equation, normally simplified, and it should pass through the stated point.
Horizontal and vertical radii produce vertical and horizontal tangents respectively.

The tangent at a point on a circle is perpendicular to the radius.

Worked example

The point $P(5,12)$ lies on $x^2+y^2=169$ . Find the equation of the tangent at $P$ .

1.The radius $OP$ has gradient $12/5$ , so the tangent gradient is $-5/12$ .
2.Use point-gradient form: $y-12=-\dfrac{5}{12}(x-5)$ .
3.Multiply by $12$ and rearrange to obtain $5x+12y=169$ .

Answer: $5x+12y=169$ .

Common mistakes

Don't use the radius gradient $12/5$ as the tangent gradient.
Don't find the negative reciprocal correctly but write a line that does not pass through the given point.

Exam tip

Substitute the contact point into your final tangent equation for a quick accuracy check.

Worked practice

Tier 1 · Easy

Write down the equation of the circle with centre

(0,0)

and radius

7

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$x^2+y^2=49$	1	Use $x^2+y^2=r^2$ with $r=7$ . Since $7^2=49$ , the equation is $x^2+y^2=49$ .

Tier 2 · Standard

The point

P(5,12)

lies on the circle

x^2+y^2=169

. Determine the tangent's equation there.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$5x+12y=169$	4	The radius $OP$ has gradient $12/5$ , so the tangent gradient is $-5/12$ . Using point-gradient form gives $y-12=-\frac{5}{12}(x-5)$ . Multiplying by $12$ and rearranging gives $5x+12y=169$ .

Tier 3 · Hard

The tangent to

x^2+y^2=50

Q(1,7)

meets the positive coordinate axes. Find the exact area of the triangle enclosed by the tangent and the axes.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$\frac{1250}{7}$ square units	5	The tangent at $(1,7)$ is $x+7y=50$ . Its intercepts are $(50,0)$ and $(0,50/7)$ . The enclosed right triangle therefore has area $\frac12\times50\times\frac{50}{7}=\frac{1250}{7}$ square units.

Tier 1 · Easy

Write down the radius of the circle

x^2+y^2=196

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$14$	1	For a circle centred at the origin, $x^2+y^2=r^2$ . Therefore $r=\sqrt{196}=14$ .

Tier 2 · Standard

Find an equation of the tangent to

x^2+y^2=81

(0,-9)

(2)

(Total for Question 5 is 2 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$y=-9$	2	The radius from $(0,0)$ to $(0,-9)$ is vertical. The tangent is perpendicular to it, so it is the horizontal line through $(0,-9)$ , namely $y=-9$ .

Tier 3 · Hard

The line

8x+15y=289

is tangent to a circle whose centre is

(0,0)

. Work out the coordinates of the point of contact.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$(8,15)$	4	The tangent has gradient $-8/15$ , so the perpendicular radius has gradient $15/8$ and equation $y=15x/8$ . Substituting into the tangent gives $8x+15(15x/8)=289$ . Multiplying by $8$ gives $64x+225x=2312$ , so $289x=2312$ and $x=8$ . Then $y=15$ , giving the point $(8,15)$ .

Tier 2 · Standard

The point

(a,12)

lies on the circle

x^2+y^2=225

. Given that

a<0

, work out the value of

a

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$a=-9$	3	Substitute $(a,12)$ into the circle equation: $a^2+12^2=225$ . Hence $a^2=81$ , so $a=9$ or $a=-9$ . The condition $a<0$ gives $a=-9$ .

Tier 3 · Hard

The point

P(-12,5)

lies on the circle

x^2+y^2=169

. Work out an equation of the tangent at

P

. The tangent meets the line

y=x

Q

. Work out the exact coordinates of

Q

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	The tangent is $-12x+5y=169$ (or $y=\dfrac{12}{5}x+\dfrac{169}{5}$ ) $Q\left(-\dfrac{169}{7},-\dfrac{169}{7}\right)$	5	The radius $OP$ has gradient $-5/12$ , so the tangent gradient is $12/5$ . An equation through $(-12,5)$ is $y-5=\dfrac{12}{5}(x+12)$ , which rearranges to $-12x+5y=169$ . At $Q$ , $y=x$ , so $-12x+5x=169$ . Hence $x=-169/7$ and $y=-169/7$ .

Tier 3 · Hard

The points

P(6,8)

and

Q(8,6)

lie on the circle

x^2+y^2=100

. Find an equation of the tangent at each point. Hence work out the exact coordinates of the point where the two tangents meet.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	At $P$ : $3x+4y=50$ (or $y=-\tfrac34x+\tfrac{25}{2}$ ) At $Q$ : $4x+3y=50$ (or $y=-\tfrac43x+\tfrac{50}{3}$ ) $\left(\dfrac{50}{7},\dfrac{50}{7}\right)$	5	For a circle centred at the origin, the tangent at $(a,b)$ has equation $ax+by=r^2$ . The tangents are therefore $6x+8y=100$ , or $3x+4y=50$ , and $8x+6y=100$ , or $4x+3y=50$ . Subtracting the equations gives $x=y$ . Hence $7x=50$ , so both coordinates are $50/7$ .

Q10

Tier 3 · Hard

A circle has centre

(0,0)

and radius

25

. A tangent has gradient

7/24

and touches the circle at a point in the upper-left part of the coordinate plane. Work out the coordinates of the point of contact and find an equation of the tangent.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	Point of contact $=(-7,24)$ $-7x+24y=625$ (or $y=\dfrac{7}{24}x+\dfrac{625}{24}$ )	4	The radius is perpendicular to the tangent, so its gradient is $-24/7$ . A radius- $25$ direction triangle has horizontal and vertical changes $7$ and $24$ . The stated position fixes the signs uniquely as $(-7,24)$ , and $(-7)^2+24^2=625$ verifies that this coordinate lies on the circle. The tangent at $(a,b)$ to $x^2+y^2=625$ is $ax+by=625$ , giving $-7x+24y=625$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-06	1H	Q20	4	Non-calculator	Higher	QP MS
2024-06	1H	Q23	5	Non-calculator	Higher	QP MS
2022-11	2H	Q24	4	Allowed	Higher	QP MS
2023-06	3H	Q23	4	Allowed	Higher	QP MS
2024-11	1H	Q20	4	Non-calculator	Higher	QP MS
2019-06	3H	Q22	4	Allowed	Higher	QP MS
2021-11	1H	Q20	3	Non-calculator	Higher	QP MS

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