Let $C_{1}$ and $C_{2}$ be two circles whose equations are given as $x^{2} + y^{2} = 25$ and $x^{2} + y^{2} + 10 x + 6 y + 1 = 0.$ $C_{3}$ is a variable circle which cuts $C_{1}$ and $C_{2}$ orthogonally. Tangents are drawn from centre of $C_{3}$ to $C_{1}$ . If the locus of mid-point of chord of contact of tangents is $α x + 3 y + \frac{13}{β} (x^{2} + y^{2}) = 0$ , then the value of $\frac{β}{α}$ is

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Solution

$C_{1} : x^{2} + y^{2} = 25$
$C_{2} : x^{2} + y^{2} + 10 x + 6 y + 1 = 0$
Radical axis is $5 x + 3 y + 13 = 0$
and centre of $C_{3}$ is $(h, - \frac{5 h + 13}{3})$
According to question, equation of chord of contact is
$h x - (\frac{5 h + 13}{3}) y = 25 \dots (1)$
and chord whose mid point is $(u, v)$ is $T = S_{1}$
i.e., $u x + v y = u^{2} + v^{2} \dots (2)$

$(1)$ and $(2)$ are identical.
$\Rightarrow h = \frac{25 u}{u^{2} + v^{2}}$
and $\frac{5 h + 13}{3} = \frac{- 25 v}{u^{2} + v^{2}}$
Eliminating $h$ , we get
$5 u + \frac{13}{25} (u^{2} + v^{2}) + 3 v = 0$
Hence, locus of mid-point of chord of contact of tangents is
$5 x + \frac{13}{25} (x^{2} + y^{2}) + 3 y = 0$
$∴ α = 5, β = 25$
$\Rightarrow \frac{β}{α} = 5$

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