Practise Equation of a circle and tangents. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
(a) A circle has equation \(x^2 + y^2 = 121\). Write down the radius of the circle.[1]
(b) Write down the coordinates of the two points where this circle crosses the \(x\)-axis.[1]
Show the answer and mark scheme
(a)Answer: \(11\)
B1 for 11
Worked solution: \(r^2 = 121\) so \(r = 11\).
(b)Answer: \((-11, 0)\) and \((11, 0)\)
B1 for \((-11, 0)\) and \((11, 0)\)
Worked solution: The centre is the origin and the radius is 11, so the circle crosses the \(x\)-axis at \((-11, 0)\) and \((11, 0)\).
Question 2Medium4 marks
A circle has equation \(x^2 + y^2 = 65\).
(a) Does the point \((4, -7)\) lie on the circle? You must show how you get your answer.[2]
(b) The point \((p, 4)\) lies on the circle. Find the possible values of \(p\).[2]
Show the answer and mark scheme
(a)Answer: Yes: \(4^2 + (-7)^2 = 65\)
M1 for \(4^2 + (-7)^2\)
C1 for yes with 65 = 65
Worked solution: \(4^2 + (-7)^2 = 16 + 49 = 65\), which equals 65, so the point is on the circle.
(b)Answer: \(p = 7\) or \(p = -7\)
M1 for \(p^2 + 4^2 = 65\) or \(p^2 = 49\)
A1 for \(p = 7\) and \(p = -7\)
Worked solution: \(p^2 + 16 = 65\) so \(p^2 = 49\) and \(p = \pm 7\).
Question 3Hard4 marks
The point \(P(3, 4)\) lies on the circle \(x^2 + y^2 = 25\). \(O\) is the origin. Jess says, 'The gradient of the radius \(OP\) is \(\frac{4}{3}\), so the gradient of the tangent to the circle at \(P\) is also \(\frac{4}{3}\).'
(a) Explain why Jess is wrong.[1]
(b) Find an equation of the tangent to the circle at \(P\).[3]
Show the answer and mark scheme
(a)Answer: The tangent is perpendicular to the radius, so its gradient is the negative reciprocal of \(\frac{4}{3}\), which is \(-\frac{3}{4}\).
C1 for stating that the tangent is perpendicular to the radius at \(P\), so its gradient is \(-\frac{3}{4}\) (not the same as the radius)
Worked solution: A tangent meets the radius at 90°. Perpendicular gradients multiply to −1, so the tangent's gradient is \(-\frac{3}{4}\).