### (i)
Centre (1, –5), $(x-1)^2 + (y+5)^2 - 19 - 1 - 25 = 0$, $(x-1)^2 + (y+5)^2 = 45$, Radius $= \sqrt{45}$

| B1 | Correct centre | |
| M1 | Correct method to find $r^2$ | $r^2 = (\pm 5)^2 + (\pm 1)^2 + 19$ for the M mark |
| A1 | Correct radius. Do not allow if wrong centre used in calculation of radius. | A0 if $\pm\sqrt{45}$ |
| [3] | | |

### (ii)
$7^2 + (-2)^2 - 14 - 20 - 19 = 0$

| B1 | Substitution of coordinates into equation of circle in any form or use of Pythagoras' theorem to calculate the distance of (7, -2) from C | No follow through for this part as AG. Must be consistent– do not allow finding distance as $\sqrt{45}$ if no/wrong radius found in 9(i). |
| [1] | | |

### (iii)
gradient of radius $= \frac{-5-(-2)}{1-7}$ or $\frac{-2-(-5)}{7-1} = \frac{1}{2}$, gradient of tangent $= -2$, $y + 2 = -2(x - 7)$, $2x + y - 12 = 0$

| M1 | uses $\frac{y_2 - y_1}{x_2 - x_1}$ with their C (3/4 correct) | Follow through from 9(i) until final mark. |
| A1√ | Follow through from their C, allow unsimplified single fraction e.g. $\frac{-3}{-6}$ | |
| B1√ | Follow through from their gradient, even if M0 scored. Allow $\frac{\text{their fraction}}$ B1 | |
| M1 | correct equation of straight line through (7, -2), any non-zero numerical gradient oe 3 term equation in correct form i.e. $k(2x + y - 12) = 0$ where $k$ is an integer | cao |
| A1 | | |
| [5] | | |

**Alternative markscheme for implicit differentiation:**
M1 Attempt at implicit diff as evidenced by $2y\frac{dy}{dx}$ term
A1 $2x + 2y\frac{dy}{dx} - 2 + 10\frac{dy}{dx} = 0$
A1 Substitution of (7, –2) to obtain gradient of tangent $= -2$
Then M1 A1 as main scheme