June 2018 Paper 2 Q17
17 A circle has equation \(\qquad x^2 + y^2 = 29\)
\(P\) is the point \((-5,\ 2)\)
(a) Show that \(P\) is on the circle. [1 mark]
(b) The tangent to the circle at \(P\) intersects the \(x\)-axis at point \(Q\).
Work out the \(x\)-coordinate of \(Q\).
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((-5)^2 + 2^2 = 29\) | B1 | oe involving use of \(-5\) and 2 eg \((-5 - 0)^2 + (2 - 0)^2 = 29\) or \((0 - -5)^2 + (0 - 2)^2 = 29\) or \(\sqrt{(-5)^2 + 2^2} = \sqrt{29}\) or \(29 - (-5)^2 = 2^2\) or \(29 - 2^2 = (-5)^2\) or \(\sqrt{29 - (-5)^2} = 2\) or \(\sqrt{29 - 2^2} = -5\) |
Additional guidance
| \(25 + 4 = 29\) | B0 |
| \(-5^2 + 2^2 = 29\) | B0 |
| Allow 29 to be written as \(\sqrt{29}^{\,2}\) |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Using gradients | ||
| (gradient \(OP =\)) \(\dfrac{2 - 0}{-5 - 0}\) or \(-\dfrac{2}{5}\) or \(-0.4\) | M1 | oe may be implied eg \(y = -\dfrac{2}{5}x\) or gradient of tangent \(= \dfrac{5}{2}\) (with gradient \(OP\) not seen) |
| (gradient tangent \(=\)) \(\dfrac{-1}{\text{their } -\dfrac{2}{5}}\) or \(\dfrac{5}{2}\) or 2.5 | M1 | oe correct or ft their \(-\dfrac{2}{5}\) |
| \(y - 2 = \text{their } \dfrac{5}{2}(x - -5)\) or \(0 - 2 = \text{their } \dfrac{5}{2}(x - -5)\) or \(2 = \text{their } \dfrac{5}{2} \times -5 + c\) | M1dep | oe dep on 2nd M1 equation of their tangent with or without substitution of \(y = 0\) implied by \(y = \dfrac{5}{2}x + \dfrac{29}{2}\) oe or \(0 = \dfrac{5}{2}x + \dfrac{29}{2}\) oe |
| \(-\dfrac{29}{5}\) or \(-5.8\) | A1 | oe allow \(\left(-\dfrac{29}{5},\ 0\right)\) SC2 answer \(-10\) (grad tangent \(= \dfrac{2}{5}\)) SC2 answer \(-\dfrac{21}{5}\) or \(-4.2\) oe (grad tangent \(= -\dfrac{5}{2}\)) |
| Alternative method 2 Using similar triangles (see diagram in Additional guidance) | ||
| \(\dfrac{a}{2} = \dfrac{2}{5}\) | M1 | oe equation any letter |
| \(a = \dfrac{2}{5} \times 2\) or \(a = \dfrac{4}{5}\) | M1dep | |
| \(-5 - \text{their } \dfrac{4}{5}\) | M1dep | dep on M2 |
| \(-\dfrac{29}{5}\) or \(-5.8\) | A1 | oe allow \(\left(-\dfrac{29}{5},\ 0\right)\) SC2 answer \(-10\) (grad tangent \(= \dfrac{2}{5}\)) SC2 answer \(-\dfrac{21}{5}\) or \(-4.2\) oe (grad tangent \(= -\dfrac{5}{2}\)) |
Additional guidance
| Alt 1 2nd M mark is not dependent but there must be a numerical value for grad \(OP\) to ft | |
| grad \(OP = -0.4\) and grad tangent \(= -0.4\) | M1M0M0A0 |
| \(\left(0,\ -\dfrac{29}{5}\right)\) | M3A0 |
| Ignore any incorrect conversion between fraction and decimal after correct answer seen | |
Alt 2 diagram |