Higher June 2022 Paper 2R Q20
20 The centre \(O\) of a circle has coordinates (4, 7)
The point \(A\), on the circle, has coordinates (6, 11) and \(AOP\) is a diameter of the circle.
Find an equation of the tangent to the circle at the point \(P\)
(4)
| Scheme | Marks |
|---|---|
| eg (\(x\) =) 4 – (6 – 4) (= 2) (\(y\) =) 7 – (11 – 7) (= 3) or (2, 3) | M1 |
eg \(\dfrac{11 - 7}{6 - 4}\;(= 2)\) or \(\dfrac{11 - [3]}{6 - [2]}\;(= 2)\) oe or \(\dfrac{[3] - 7}{[2] - 4}\;(= 2)\) oe | M1 |
| eg \(-1 \div [2]\;(= -0.5)\) oe | M1ft |
| \(y - 3 = -0.5(x - 2)\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for a method to find the coordinates of \(P\) (accept coordinates of \(P\) informally eg separately or as a vector)
M1: (indep if using coordinates of \(A\) & \(O\))
for a method to find the gradient of \(AOP\) (can use their coordinates of \(P\))
M1ft: for a method to find the gradient of the tangent
ft their stated gradient of \(AOP\) (or \(OA\) or \(OP\)) (could be embedded)
A1: oe eg \(y = -\dfrac{1}{2}x + 4\)