(a)     The function \(g\) is defined by \(g(x,{\text{ }}y) = {x^2} + {y^2} + dx + ey + f\) and the circle \({C_1}\) has equation \(g(x, y) = 0\).

(i)     Show that the centre of \({C_1}\) has coordinates \(\left( { - \frac{d}{2}, - \frac{e}{2}} \right)\) and the radius of \({C_1}\) is \(\sqrt {\frac{{{d^2}}}{4} + \frac{{{e^2}}}{4} - f} \).

(ii)     The point \({\text{P}}(a, b)\) lies outside \({C_1}\). Show that the length of the tangents from \({\text{P}}\) to \({C_1}\) is equal to \(\sqrt {g(a,{\text{ }}b)} \).

(b)     The circle \({C_2}\) has equation \({x^2} + {y^2} - 6x - 2y + 6 = 0\).

The line \(y = mx\) meets \({C_2}\) at the points \({\text{R}}\) and \({\text{S}}\).

(i)     Determine the quadratic equation whose roots are the x-coordinates of \({\text{R}}\) and \({\text{S}}\).

(ii)     Hence, given that \(L\) denotes the length of the tangents from the origin \({\text{O}}\) to \({C_2}\), show that \({\text{OR}} \times {\text{OS}} = {L^2}\).

(a)     (i)     completing the square,

\({\left( {x + \frac{d}{2}} \right)^2} + {\left( {y + \frac{e}{2}} \right)^2} - \frac{{{d^2}}}{4} - \frac{{{e^2}}}{4} + f = 0\)     M1A1

whence the centre \({\text{C}}\) is the point \(\left( { - \frac{d}{2}, - \frac{e}{2}} \right)\) and the radius is

\(\sqrt {\frac{{{d^2}}}{4} + \frac{{{e^2}}}{4} - f} \)     AG

(ii)     \({\text{C}}{{\text{P}}^2} = {\left( {a + \frac{d}{2}} \right)^2} + {\left( {b + \frac{e}{2}} \right)^2}\)     (A1)

let \({\text{Q}}\) denote the point of contact of one of the tangents from \({\text{P}}\) to the circle.

\({\text{C}}{{\text{Q}}^2} = \frac{{{d^2}}}{4} + \frac{{{e^2}}}{4} - f\)     (A1)

using Pythagoras’ Theorem in triangle \({\text{CPQ}}\),     M1

\({L^2} = {\left( {a + \frac{d}{2}} \right)^2} + {\left( {b + \frac{e}{2}} \right)^2} - \left( {\frac{{{d^2}}}{4} + \frac{{{e^2}}}{4} - f} \right)\)

\( = {a^2} + {b^2} + da + eb + f = g(a, b)\)     A1

therefore \(L = \sqrt {g(a,{\text{ }}b)} \)     AG

[6 marks]

(b)     (i)     the x-coordinates of \({\text{R, S}}\) satisfy

\({x^2} + {(mx)^2} - 6x - 2mx + 6 = 0\)     M1

\((1 + {m^2}){x^2} - (6 + 2m)x + 6 = 0\)     A1

(ii)     \({L^2} = g(0,{\text{ }}0) = 6\)     A1

let  \({x_1},{x_2}\) denote the two roots. Then \({x_1}{x_2} = \frac{6}{{1 + {m^2}}}\)     A1

\({\text{OR}} = \sqrt {x_1^2 + {{(m{x_1})}^2}}  = {x_1}\sqrt {1 + {m^2}} \) and \({\text{OS}} = {x_2}\sqrt {1 + {m^2}} \)     M1

therefore

\({\text{OR}} \times {\text{OS}} = {x_1}{x_2}(1 + {m^2}) = 6\)     A1

so that \({\text{OR}} \times {\text{OS}} = {L^2}\)     AG

[6 marks]