C2 January 2007 Q3
3. The line joining the points \((-1, 4)\) and \((3, 6)\) is a diameter of the circle \(C\).
Find an equation for \(C\). (6)
| Scheme | Marks |
|---|---|
| Centre \(\left(\dfrac{-1 + 3}{2}, \dfrac{6 + 4}{2}\right)\), i.e. \((1, 5)\) | M1, A1 |
| \(r = \dfrac{\sqrt{(3 - (-1))^2 + (6 - 4)^2}}{2}\) or \(r^2 = (1 - (-1))^2 + (5 - 4)^2\) or \(r^2 = (3 - 1)^2 + (6 - 5)^2\) o.e. | M1 |
| \((x - 1)^2 + (y - 5)^2 = 5\) | M1, A1, A1 |
| (6) |
Notes
M1: Some use of correct formula in \(x\) or \(y\) coordinate. Can be implied.
Use of \(\left(\tfrac{1}{2}(x_A - x_B), \tfrac{1}{2}(y_A - y_B)\right) \to (-2, -1)\) or \((2, 1)\) is M0 A0 but watch out for use of \(x_A + \tfrac{1}{2}(x_A - x_B)\) etc which is okay.
A1: \((1, 5)\)
\((5, 1)\) gains M1 A0.
M1: Correct method to find \(r\) or \(r^2\) using given points or f.t. from their centre. Does not need to be simplified.
Attempting radius \(= \sqrt{\dfrac{(\text{diameter})^2}{2}}\) is an incorrect method, so M0.
N.B. Be careful of labelling: candidates may not use \(d\) for diameter and \(r\) for radius.
Labelling should be ignored.
Simplification may be incorrect – mark awarded for correct method.
Use of \(\sqrt{(x_1 - x_2)^2 - (y_1 - y_2)^2}\) is M0.
M1: Write down \((x \pm a)^2 + (y \pm b)^2 =\) any constant (a letter or a number).
Numbers do not have to be substituted for \(a\), \(b\) and if they are they can be wrong.
A1: LHS is \((x - 1)^2 + (y - 5)^2\). Ignore RHS.
A1: RHS is 5.
Ignore subsequent working. Condone use of decimals that leads to exact 5.
Or correct equivalents, e.g. \(x^2 + y^2 - 2x - 10y + 21 = 0\).
Alternative
Alternative – note the order of the marks needed for ePEN.
| As above. | M1 |
| As above. | A1 |
| \(x^2 + y^2 + (\text{constant})x + (\text{constant})y + \text{constant} = 0\). Numbers do not have to be substituted for the constants and if they are they can be wrong. | 3rd M1 |
| Attempt an appropriate substitution of the coordinates of their centre (i.e. working with coefficient of \(x\) and coefficient of \(y\) in equation of circle) and substitute \((-1, 4)\) or \((3, 6)\) into equation of circle. | 2nd M1 |
| \(-2x - 10y\) part of the equation \(x^2 + y^2 - 2x - 10y + 21 = 0\). | A1 |
| \(+21 = 0\) part of the equation \(x^2 + y^2 - 2x - 10y + 21 = 0\). | A1 |
| Or correct equivalents, e.g. \((x - 1)^2 + (y - 5)^2 = 5\). |