June 2019 Paper 2 Q23
23 A circle has centre \(C\) and equation \(\quad (x - 1)^2 + (y + 3)^2 = 25\)
\(P\,(4,\ -7)\) and \(Q\) are points on the circle.
The tangent at \(Q\) is parallel to the \(x\)-axis.
The tangents at \(P\) and \(Q\) intersect at point \(R\).
(a) Write down the coordinates of \(C\). [1 mark]
(b) Show that the equation of the tangent at \(Q\) is \(\quad y = 2\) [1 mark]
(c) Work out the \(x\)-coordinate of \(R\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((1,\ -3)\) | B1 |
Additional guidance
| Mark intention eg condone 1, −3 | B1 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(-3 + \sqrt{25}\ (= 2)\) or \(-3 + 5\ (= 2)\) | B1 | oe eg \(5 - 3\ (= 2)\) or \(2 + 3 = 5\) |
| Alternative method 2 | ||
| \((y + 3)^2 = 25\) and \(y = 2\) or \(y + 3 = 5\) and \(y = 2\) or \((2 + 3)^2 = 25\) | B1 | oe eg \((1 - 1)^2 + (y + 3)^2 = 25\) and \(y = 2\) |
Additional guidance
| \((1,\ -3) + (0,\ 5) = (1,\ 2)\) so \(y = 2\) | B0 |
| Allow \(-3\) + radius of 5 | B1 |
| \(2 = 0x + \mathrm{c}\) \(c = 2\) so \(y = 2\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Using equation \(PR\) | ||
| \(\dfrac{-7 - \text{their } -3}{4 - \text{their } 1}\) or \(-\dfrac{4}{3}\) | M1 | oe grad \(PC\) their \(-3\) and their 1 from (a) |
| \(-1 \div \text{their } -\dfrac{4}{3}\) or \(\dfrac{3}{4}\) | M1 | oe grad \(PR\) their \(-\dfrac{4}{3}\) must be a value (gradient \(PR =\)) \(\dfrac{3}{4}\) is M2 |
| \(2 - -7 = \text{their } \dfrac{3}{4}(x - 4)\) | M1dep | oe equation \(PR\) with \(y = 2\) substituted eg \(2 = \dfrac{3}{4}x - 10\) dep on 2nd M1 |
| 16 | A1ft | only ft their \(-3\) and their 1 from (a) |
| Alternative method 2 Using \(RC^2 = CP^2 + PR^2\) or \(PR^2 = QR^2\) with \(R\,(x,\ 2)\) | ||
| \((x - \text{their } 1)^2 + (2 - \text{their } -3)^2\) \(= (2 - \text{their } -3)^2 + (x - 4)^2 + (2 - -7)^2\) | M1 | oe eg \((x - 1)^2 = (x - 4)^2 + (2 - -7)^2\) their \(-3\) and their 1 from (a) |
| \(x^2 - 2x + 1 + 25\) \(= 25 + x^2 - 8x + 16 + 81\) | M1dep | oe brackets expanded |
| \(96 = 6x\) or \(96 \div 6\) | M1dep | oe linear equation or calculation dep on M2 |
| 16 | A1ft | only ft their \(-3\) and their 1 from (a) |
| Alternative method 3 Using equation \(CR\) | ||
| \(\dfrac{-7 - 2}{4 - \text{their } 1}\) or \(-3\) | M1 | oe grad \(PQ\) their 1 from (a) |
| \(-1 \div \text{their } -3\) or \(\dfrac{1}{3}\) | M1 | oe grad \(CR\) their \(-3\) must be a value (gradient \(CR =\)) \(\dfrac{1}{3}\) is M2 |
| \(2 - \text{their } -3 = \text{their } \dfrac{1}{3}(x - \text{their } 1)\) | M1dep | oe equation \(CR\) with \(y = 2\) substituted eg \(2 = \dfrac{1}{3}x - \dfrac{10}{3}\) dep on 2nd M1 |
| 16 | A1ft | only ft their \(-3\) and their 1 from (a) |
| Alternative method 4 Using equation \(MR\) where \(M\) is the midpoint of \(PQ\) | ||
| \(\dfrac{-7 - 2}{4 - \text{their } 1}\) or \(-3\) | M1 | oe grad \(PQ\) their 1 from (a) |
| \(-1 \div \text{their } -3\) or \(\dfrac{1}{3}\) | M1 | oe grad \(MR\) their \(-3\) must be a value (gradient \(MR =\)) \(\dfrac{1}{3}\) is M2 |
| \(\left(\dfrac{4 + \text{their } 1}{2},\ \dfrac{-7 + 2}{2}\right)\) or \((2.5,\ -2.5)\) and \(2 - \text{their } -2.5 = \text{their } \dfrac{1}{3}(x - \text{their } 2.5)\) | M1dep | oe midpoint of \(PQ\) and equation \(MR\) with \(y = 2\) substituted eg \(2 = \dfrac{1}{3}x - \dfrac{10}{3}\) dep on 2nd M1 |
| 16 | A1ft | only ft their 1 from (a) |
| Alternative method 5 Using equation \(MC\) where \(M\) is the midpoint of \(PQ\) | ||
| \(\left(\dfrac{4 + \text{their } 1}{2},\ \dfrac{-7 + 2}{2}\right)\) or \((2.5,\ -2.5)\) | M1 | oe midpoint of \(PQ\) their 1 from (a) |
| \(\dfrac{\text{their } -3 - \text{their } -2.5}{\text{their } 1 - \text{their } 2.5}\) or \(\dfrac{1}{3}\) | M1dep | oe grad \(MC\) |
| \(2 - \text{their } -3 = \text{their } \dfrac{1}{3}(x - \text{their } 1)\) or \(2 - \text{their } -2.5 = \text{their } \dfrac{1}{3}(x - \text{their } 2.5)\) | M1dep | oe equation \(MC\) with \(y = 2\) substituted eg \(2 = \dfrac{1}{3}x - \dfrac{10}{3}\) dep on M2 |
| 16 | A1ft | only ft their \(-3\) and their 1 from (a) |
| Alternative method 6 Using trigonometry where \(M\) is the midpoint of \(PQ\) | ||
| \((QM =)\ \dfrac{1}{2}\sqrt{(4 - \text{their } 1)^2 + (-7 - 2)^2}\) or \(\dfrac{1}{2}\sqrt{90}\) or 4.74… | M1 | |
| \(\sin^{-1}\left(\dfrac{\text{their } 4.74\ldots}{5}\right)\) or (angle \(QCM =\)) 71.5… or 71.6 | M1dep | oe angle \(QCM\) |
| \(\tan(\text{their } 71.5\ldots) = \dfrac{x - \text{their } 1}{5}\) | M1dep | using triangle \(QCR\) |
| 16 | A1ft | only ft their 1 from (a) |
Additional guidance
| Allow (16, …) to imply answer 16 | |
| Alt 1 \(-\dfrac{4}{3}x\) is M0 unless recovered | |
| (a) \((1,\ -2)\) grad \(PC = -\dfrac{5}{3} \qquad\) grad \(PR = \dfrac{3}{5}\) Answer 19 (3rd M1 can be implied by A1ft answer) | M1M1 M1A1ft |