CAIE IGCSE Additional Math 8 coordinate geometry of the circle

$$(3-4)^{2}+(-1+3)^{2}=5$$

or showing that distance between point A and centre of circle = radius e.g.
$$\sqrt{(3-4)^{2}+(-1+3)^{2}}=\sqrt{5}$$

B1

Accept if x-coordinate substituted to find
y - coordinate

gradient of radius $=\frac{-3+1}{4-3}=-2$ soi by gradient
of tangent

M1

gradient of tangent $=\frac{-1}{\text { their }(-2)}$

M1

FT their -2
Allow if differentiation is used e.g.:

$y+1=\frac{1}{2}(x-3)$ oe

A1

$$\text { FT } \frac{-1}{\text { their }-2}$$

ISW from a correct unsimplified answer

Alternative

use of differentiation e.g.:
$$\begin{aligned} (y+3)^{2}=5-(x-4)^{2} y=\sqrt{-x^{2}+8 x-11}-3 \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-2 x+8}{2 \sqrt{-x^{2}+8 x-11}} \end{aligned}$$

(M1)

allow one error
or use of implicit differentiation (not on syllabus) e.g.:
$2 x+2 y \frac{\mathrm{~d} y}{\mathrm{~d} x}-8+6 \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
leading to $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{8-2 x}{2 y+6}$

Substitute (3,-1) in their $\frac{\mathrm{d} y}{\mathrm{~d} x}$
to get gradient $=\frac{1}{2}$

(M1)

Dep on first M1

$y+1=\frac{1}{2}(x-3)$ oe

(A1)

ISW from a correct unsimplified answer

$(x-4)^2+(y-2)^2=40$ oe, isw
or
$x^2+y^2-8x-4y-20=0$ oe, isw
B2
B1 for $r^2=40$ or $r=\sqrt{40}$ oe, soi.

$16+(y-2)^2=40$ oe
M1 FT their equation of the circle, providing it is the equation of a circle.
$(y-2)^2=24$ or $y^2-4y-20=0$
A1 for correct quadratic equation in any form.
$(0,2\pm\sqrt{24})$ isw or equivalent exact form
A1