(i)     the value of $k$;

(ii)     the radius of $C$;

(iii)     the $x$-intercepts of $C$.

Let M be any point on $C$ and N be the $x$-intercept of $C$ between A and B.

Prove that ${\rm{A\hat MN}} = {\rm{N\hat MB}}$.

(i)     let $(x,{\text{ }}y)$ be a point on $C$

then ${(x + 3)^2} + {y^2} = {k^2}\left( {{{(x - 5)}^2} + {y^2}} \right)$     M1A1A1

Note:     Award M1 for form of an Apollonius circle, A1 for each side.

rearrange, for example,

$({k^2} - 1){x^2} - (10{k^2} + 6)x + ({k^2} - 1){y^2} + 25{k^2} - 9 = 0$    A1

equate the $x$-coordinate of the centre as given by this equation to 13:

$\frac{{5{k^2} + 3}}{{{k^2} - 1}} = 13$    M1A1

obtain ${k^2} = 2 \Rightarrow k = \sqrt 2$     A1

(ii)     METHOD 1

with this value of $k$, the equation can be reduced to the form

${(x - 13)^2} + {y^2} = 128$    M1A1

obtain the radius $\sqrt {128} {\text{ }}\left( { = 8\sqrt 2 } \right)$     A1

METHOD 2

assuming N is the $x$-intercept of $C$ between A and B

$\frac{{{\text{AB}}}}{{{\text{BN}}}} = \frac{{16 - r}}{{r - 8}} = \sqrt 2$    M1A1

$\Rightarrow r = 8\sqrt 2$    A1

Note:     Accept answers given in terms of $k$, if no value of $k$ found in (a)(i).

(iii)     $x$-intercepts are $13 \pm 8\sqrt 2$     A1

[11 marks]

because N lies on the circle it satisfies the Apollonius property

hence ${\text{AN}} = \sqrt 2 {\text{NB}}$     R1

but as ${\text{AM}} = \sqrt 2 {\text{MB}}$     R1

by the converse to the angle-bisector theorem     R1

${\rm{A\hat MN}} = {\rm{N\hat MB}}$    AG

[3 marks]

This question was the first one on the paper to cause a significant problem for the majority of candidates. Many were unable to start and a small number were unable to successfully deal with the algebraic manipulation required from the method they had embarked upon. For those who were successful at part (a), part (b) was often not fully correct, again due to the degree of formality required from the command term “prove”.

This question was the first one on the paper to cause a significant problem for the majority of candidates. Many were unable to start and a small number were unable to successfully deal with the algebraic manipulation required from the method they had embarked upon. For those who were successful at part (a), part (b) was often not fully correct, again due to the degree of formality required from the command term “prove”.