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Equation of a Circle

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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.

Have a pen and paper handy, to work out your answers.

Question 12 marks

A circle has centre (-4, 3) and touches the y-axis.

Work out the equation of the circle.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

If the circle just touches the y-axis, the radius is the distance from the centre to the y-axis.

Worked solution
  1. The distance from (-4, 3) to the y-axis is 4 (the size of the x-coordinate).
  2. The circle touches the y-axis, so the radius is 4 (not 3, which would be the distance to the x-axis).
  3. (x-a)^2+(y-b)^2=r^2 with a=-4, b=3, r=4
  4. (x+4)^2+(y-3)^2=4^2
  5. Answer: (x+4)^2+(y-3)^2=16

Question 22 marks

The point P(1, 8) lies on the circle (x+1)^2+(y-2)^2=40 Work out the gradient of the tangent to the circle at P.

Give your answer as a fraction.

Hint

The tangent at P is perpendicular to the radius from the centre to P.

Worked solution
  1. The centre is (-1, 2)
  2. Gradient of the radius from (-1, 2) to P(1, 8): \dfrac{8-2}{1-(-1)}=\dfrac62=3
  3. The tangent is perpendicular to the radius, so its gradient is the negative reciprocal of 3.
  4. Answer: -\dfrac13

Question 3Challenge6 marks

A and B are two points on the circle x^2+y^2=130 The midpoint of the chord AB is M(4, -4).

(a)

Work out the equation of the line AB.

Give your answer in the form y=mx+c

3 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

(b)

Work out the length of AB.

Give your answer in the form k\sqrt{2}, where k is an integer.

3 marks

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

Join the centre O to M: the radius through the midpoint of a chord meets the chord at right angles.

Worked solution

Part (a)

  1. The centre of the circle is O(0, 0)
  2. The line from the centre to the midpoint of a chord is perpendicular to the chord.
  3. Gradient of OM=\dfrac{-4-0}{4-0}=-1
  4. So the gradient of AB is the negative reciprocal: 1
  5. AB passes through M(4, -4): y-(-4)=1(x-4)
  6. Answer: y=x-8

Part (b)

  1. Method 1 (Pythagoras): OM^2=4^2+4^2=32 and OA^2=r^2=130
  2. Triangle OMA has a right angle at M, so MA^2=130-32=98 and MA=\sqrt{98}=7\sqrt2
  3. M is the midpoint, so AB=2\times7\sqrt2=14\sqrt2
  4. Method 2 (intersection): substitute y=x-8 into x^2+y^2=130: 2x^2-16x+64=130
  5. x^2-8x-33=0, so (x-11)(x+3)=0, giving A(11, 3) and B(-3, -11)
  6. AB=\sqrt{14^2+14^2}=\sqrt{392}=14\sqrt2
  7. Answer: 14\sqrt2

More on this topic: Equation of a Circle worksheet with full solutions

All AQA Level 2 Further Maths practice questions

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