November 2021 Paper 1 Q12
12 A circle, centre \(C\ (4, -2)\), passes through the origin and point \(A\ (8, 0)\) on the \(x\)-axis.
The tangent at \(A\) is shown.
Not drawn accurately
(a) Work out the equation of the circle. [2 marks]
(b) Work out the equation of the tangent to the circle at \(A\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((x - 4)^2 + (y + 2)^2 = 20\) | B2 | B1 \((x - 4)^2 + (y + 2)^2\) or 20 |
Additional guidance
| \((x + 4)^2 + (y - 2)^2 = 20\) \((x - 4)^2 + (y + 2)^2 = 4^2 + (-2)^2\) \((x - 4)^2 + (y + 2)^2 = \sqrt{20}\) \((x - 4)^2 - (y + 2)^2 = 20\) \((x - 4)^2 + (y - -2)^2 = 20\) \((x - 4)^2 + (y - -2)^2 = (\sqrt{20})^2\) ignore further working | B1 B1 B1 B1 B2 B2 |
| Answer | Mark | Comments |
|---|---|---|
| (Gradient \(AC\) =) \(\dfrac{0 - -2}{8 - 4}\) or \(\dfrac{2}{4}\) | M1 | oe |
| (Gradient of tangent =) negative reciprocal of their \(\dfrac{2}{4}\) or \(-2\) | M1 | oe ft their gradient \(AC\) only gradient \(-2\) seen is M2 |
| \(y = -2x + 16\) | A1 | oe |
Additional guidance
| It is possible to find an incorrect gradient of \(AC\) and then get the second M mark for finding the negative reciprocal of this | M0M1A0 |