C2 January 2013 Q5
5. The circle \(C\) has equation\[x^2 + y^2 - 20x - 24y + 195 = 0\]The centre of \(C\) is at the point \(M\).
\(N\) is the point with coordinates \((25, 32)\).
The tangent to \(C\) at a point \(P\) on the circle passes through point \(N\).
| Scheme | Marks |
|---|---|
| (i) The centre is at \((10, 12)\) | B1 B1 |
| (ii) Uses \((x - 10)^2 + (y - 12)^2 = -195 + 100 + 144 \Rightarrow r = \ldots\) | M1 |
| \(r = \sqrt{10^2 + 12^2 - 195}\) | A1 |
| \(r = 7\) | A1 |
| (5) |
Notes
B1: \(x = 10\) B1: \(y = 12\)
M1: Completes the square for both \(x\) and \(y\) in an attempt to find \(r\).
\((x \pm \text{"}10\text{"})^2 \pm a\) and \((y \pm \text{"}12\text{"})^2 \pm b\) and \(+195 = 0,\ (a, b \neq 0)\)
Allow errors in obtaining their \(r^2\) but must find square root
A1: A correct numerical expression for \(r\) including the square root and can implied by a correct value for \(r\)
A1: Not \(r = \pm 7\) unless \(-7\) is rejected
(a) Way 2
| Scheme | Marks |
|---|---|
| Compares the given equation with \(x^2 + y^2 + 2gx + 2fy + c = 0\) to write down centre \((-g, -f)\) i.e. \((10, 12)\) | B1B1 |
| Uses \(r = \sqrt{(\pm\text{"}10\text{"})^2 + (\pm\text{"}12\text{"})^2 - c}\) | M1 |
| \(r = \sqrt{10^2 + 12^2 - 195}\) | A1 |
| \(r = 7\) | A1 |
| (5) |
B1: \(x = 10\) B1: \(y = 12\)
A1: A correct numerical expression for \(r\)
| Scheme | Marks |
|---|---|
| \(MN = \sqrt{(25 - \text{"}10\text{"})^2 + (32 - \text{"}12\text{"})^2}\) | M1 |
| \(MN\left(= \sqrt{625}\right) = 25\) | A1 |
| (2) |
Notes
M1: Correct use of Pythagoras
| Scheme | Marks |
|---|---|
| \(NP = \sqrt{\left(\text{"}25\text{"}^2 - \text{"}7\text{"}^2\right)}\) \(NP = \sqrt{\left(MN^2 - r^2\right)}\) | M1 |
| \(NP\left(= \sqrt{576}\right) = 24\) | A1 |
| (2) |
(c) Way 2
| Scheme | Marks |
|---|---|
| \(\cos(NMP) = \dfrac{7}{\text{"}25\text{"}} \Rightarrow NP = \text{"}25\text{"}\sin(NMP)\) | M1 |
| \(NP = 24\) | A1 |
| (2) | |
| [9] |
M1: Correct strategy for finding \(NP\)