C2 June 2009 Q6
6. The circle \(C\) has equation\[x^2 + y^2 - 6x + 4y = 12\]
The point \(P(-1,\ 1)\) and the point \(Q(7,\ -5)\) both lie on \(C\).
The point \(R\) lies on the positive \(y\)-axis and the angle \(PRQ = 90^\circ\).
| Scheme | Marks |
|---|---|
| \((x - 3)^2 - 9 + (y + 2)^2 - 4 = 12\qquad\) Centre is \((3,\ -2)\) | M1 A1, A1 |
| \((x - 3)^2 + (y + 2)^2 = 12 + \text{"}9\text{"} + \text{"}4\text{"}\qquad r = \sqrt{12 + \text{"}9\text{"} + \text{"}4\text{"}} = 5\) (or \(\sqrt{25}\) ) | M1 A1 |
| (5) |
Notes
1st M1 for attempt to complete square. Allow \((x \pm 3)^2 \pm k\), or \((y \pm 2)^2 \pm k\), \(k \neq 0\).
1st A1 \(x\)-coordinate 3, 2nd A1 \(y\)-coordinate \(-2\)
2nd M1 for a full method leading to \(r = \ldots\), with their 9 and their 4, 3rd A1 5 or \(\sqrt{25}\)
The 1st M can be implied by \((\pm 3,\ \pm 2)\) but a full method must be seen for the 2nd M.
Where the 'diameter' in part (b) has clearly been used to answer part (a), no marks in (a), but in this case the M1 (not the A1) for part (b) can be given for work seen in (a).
Alternative
1st M1 for comparing with \(x^2 + y^2 + 2gx + 2fy + c = 0\) to write down centre \((-g,\ -f)\) directly. Condone sign errors for this M mark.
2nd M1 for using \(r = \sqrt{g^2 + f^2 - c}\) . Condone sign errors for this M mark.
| Scheme | Marks |
|---|---|
| \(PQ = \sqrt{(7 - {-1})^2 + (-5 - 1)^2}\) or \(\sqrt{8^2 + 6^2}\) | M1 |
| \(= 10 = 2 \times \text{radius},\ \therefore \text{diam.}\) (N.B. For A1, need a comment or conclusion) | A1 |
| (2) |
Notes
[ALT: midpt. of \(PQ\ \left(\frac{7 + (-1)}{2},\frac{1 + (-5)}{2}\right)\): M1, \(= (3,\ -2) =\) centre: A1]
[ALT: eqn. of \(PQ\ \ 3x + 4y - 1 = 0\): M1, verify \((3,\ -2)\) lies on this: A1]
[ALT: find two grads, e.g. \(PQ\) and \(P\) to centre: M1, equal \(\therefore\) diameter: A1]
[ALT: show that point \(S(-1,\ -5)\) or \((7,\ 1)\) lies on circle: M1
because \(\angle PSQ = 90^\circ\), semicircle \(\therefore\) diameter: A1]
| Scheme | Marks |
|---|---|
| \(R\) must lie on the circle (angle in a semicircle theorem)… often implied by a diagram with \(R\) on the circle or by subsequent working) | B1 |
| \(x = 0 \Rightarrow\quad y^2 + 4y - 12 = 0\) | M1 |
| \((y - 2)(y + 6) = 0\quad y = \ldots..\) (M is dependent on previous M) | dM1 |
| \(y = -6\) or 2 (Ignore \(y = -6\) if seen, and 'coordinates' are not required)) | A1 |
| (4) | |
| [11] |
Notes
1st M1 for setting \(x = 0\) and getting a 3TQ in \(y\) by using eqn. of circle.
2nd M1 (dep.) for attempt to solve a 3TQ leading to at least one solution for \(y\).
Alternative 1: (Requires the B mark as in the main scheme)
1st M for using (3, 4, 5) triangle with vertices \((3,-2),(0,-2),(0,\ y)\) to get a linear or quadratic equation in \(y\) (e.g. \(3^2 + (y + 2)^2 = 25\) ).
2nd M (dep.) as in main scheme, but may be scored by simply solving a linear equation.
Alternative 2: (Not requiring realisation that \(R\) is on the circle)
B1 for attempt at \(m_{PR} \times m_{QR} = -1\), (NOT \(m_{PQ}\)) or for attempt at Pythag. in triangle \(PQR\).
1st M1 for setting \(x = 0\), i.e. (0, \(y\)), and proceeding to get a 3TQ in \(y\). Then main scheme.
Alternative 2 by 'verification':
B1 for attempt at \(m_{PR} \times m_{QR} = -1\), (NOT \(m_{PQ}\)) or for attempt at Pythag. in triangle \(PQR\).
1st M1 for trying (0, 2).
2nd M1 (dep.) for performing all required calculations.
A1 for fully correct working and conclusion.