Higher June 2023 Paper 3 Q23
23 A circle has equation \(x^2 + y^2 = 25\)
The point \(P\) with coordinates (−3, 4) lies on the circle.
Alex says that the tangent to the circle at \(P\) crosses the \(x\)-axis at the point (−8, 0)
Is Alex correct?
You must show how you get your answer. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| No (supported) | P1 | for a first step, eg to find the gradient of the line (normal) joining (0, 0) to (−3, 4), eg \(\dfrac{4 - 0}{-3 - 0}\ \left(= -\dfrac{4}{3}\right)\) OR finds the length2 for one side of triangle, eg \((-3 - -8)^2 + (4 - 0)^2\ (= 41)\) or \((-3 - 0)^2 + (4 - 0)^2\ (= 25)\) or 64 |
| P1 | for a second step, eg to find the gradient of the tangent, eg \(-1 \div \text{``}\dfrac{-4}{3}\text{''}\ \left(= \dfrac{3}{4}\right)\) OR finds the length2 for two sides of triangle, two of 25, 41 and 64 | |
| P1 | for a third step, eg to find the gradient of the line joining (−8, 0) to (−3, 4), \(\dfrac{4 - 0}{-3 - -8}\ \left(= \dfrac{4}{5}\right)\) or finds the equation of the tangent, eg \(y = \dfrac{3}{4}x + \dfrac{25}{4}\) or \(y - 4 = \dfrac{3}{4}(x - -3)\) OR for process to use Pythagoras’ rule, eg \(41 + 25\ (= 66)\) or \(64 - 25\ (= 39)\) or \(64 - 41\ (= 23)\) | |
| C1 | for No from correct figures and a complete process from comparison of gradients, eg \(\dfrac{3}{4}\) and \(\dfrac{4}{5}\) or showing the equation of the tangent does not pass through (−8, 0), eg when \(x = -8\), \(y = \dfrac{1}{4}\) (not 0) or when \(y = 0\), \(x = -\dfrac{25}{3}\) (not −8) OR correct figures from Pythagoras’ rule, eg \(41 + 25\ (= 66) \neq 64\) or \(64 - 41 \neq 25\) or \(64 - 25 \neq 41\) oe |
Additional guidance
Alternative processes may be seen and should be duly credited.
Award 0 marks for No without complete and correct supportive working