C2 January 2012 Q2
2. A circle \(C\) has centre \((-1, 7)\) and passes through the point \((0, 0)\). Find an equation for \(C\). (4)
| Scheme | Marks |
|---|---|
| The equation of the circle is \((x + 1)^2 + (y - 7)^2 = (r^2)\) | M1 A1 |
| The radius of the circle is \(\sqrt{(-1)^2 + 7^2} = \sqrt{50}\) or \(5\sqrt{2}\) or \(r^2 = 50\) | M1 |
| So \((x + 1)^2 + (y - 7)^2 = 50\) or equivalent | A1 |
| (4) | |
| 4 |
Notes
M1 is for this expression on left hand side– allow errors in sign of 1 and 7.
A1 correct signs (just LHS)
M1 is for Pythagoras or substitution into equation of circle to give \(r\) or \(r^2\)
Giving this value as diameter is M0
A1, cao for cartesian equation with numerical values but allow \(\left(\sqrt{50}\right)^2\) or \(\left(5\sqrt{2}\right)^2\) or any exact equivalent
A correct answer implies a correct method – so answer given with no working earns all four marks for this question.
Alternative method
| Scheme | Marks |
|---|---|
| Equation of circle is \(x^2 + y^2 \pm 2x \pm 14y + c = 0\) | M1 |
| Equation of circle is \(x^2 + y^2 + 2x - 14y + c = 0\) | A1 |
| Uses (0,0) to give \(c = 0\), or finds \(r = \sqrt{(-1)^2 + 7^2} = \sqrt{50}\) or \(5\sqrt{2}\) or \(r^2 = 50\) | M1 |
| So \(x^2 + y^2 + 2x - 14y = 0\) or equivalent | A1 |