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The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4 x-$ $5 y=20$ to the circle $x^{2}+y^{2}=9$ is
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The correct answer is:
$20\left(x^{2}+y^{2}\right)-36 x+45 y=0$
Any point $P$ on line $4 x-5 y=20$ can be considered as $\left(\alpha, \frac{4 \alpha-20}{5}\right)$.
Equation of chord of contact $A B$ to the circle $x^{2}+y^{2}=9$
drawn from point $\mathrm{P}\left(\alpha, \frac{4 \alpha-20}{5}\right)$ is
x. $\alpha+$ y. $\left(\frac{4 \alpha-20}{5}\right)=9...(i)$

Also the equation of chord $A B$ whose mid point is $(h, k)$ is $h x+k y=h^{2}+k^{2}....(ii)$
$\because \quad$ Equations (i) and (ii) represent the same line,
$\therefore \quad \frac{h}{\alpha}=\frac{k}{\frac{4 \alpha-20}{5}}=\frac{h^{2}+k^{2}}{9}$
$\Rightarrow 5 k \alpha=4 h \alpha-20 h$ and $9 h=\alpha\left(h^{2}+k^{2}\right)$
$\Rightarrow \alpha=\frac{20 h}{4 h-5 k} \quad$ and $\alpha=\frac{9 h}{h^{2}+k^{2}}$
$\Rightarrow \quad \frac{20 h}{4 h-5 k}=\frac{9 h}{h^{2}+k^{2}} \Rightarrow 20\left(h^{2}+k^{2}\right)=9(4 h-5 k)$
$\therefore \quad$ Locus of $(h, k)$ is $20\left(x^{2}+y^{2}\right)-36 x+45 y=0$
Equation of chord of contact $A B$ to the circle $x^{2}+y^{2}=9$
drawn from point $\mathrm{P}\left(\alpha, \frac{4 \alpha-20}{5}\right)$ is
x. $\alpha+$ y. $\left(\frac{4 \alpha-20}{5}\right)=9...(i)$

Also the equation of chord $A B$ whose mid point is $(h, k)$ is $h x+k y=h^{2}+k^{2}....(ii)$
$\because \quad$ Equations (i) and (ii) represent the same line,
$\therefore \quad \frac{h}{\alpha}=\frac{k}{\frac{4 \alpha-20}{5}}=\frac{h^{2}+k^{2}}{9}$
$\Rightarrow 5 k \alpha=4 h \alpha-20 h$ and $9 h=\alpha\left(h^{2}+k^{2}\right)$
$\Rightarrow \alpha=\frac{20 h}{4 h-5 k} \quad$ and $\alpha=\frac{9 h}{h^{2}+k^{2}}$
$\Rightarrow \quad \frac{20 h}{4 h-5 k}=\frac{9 h}{h^{2}+k^{2}} \Rightarrow 20\left(h^{2}+k^{2}\right)=9(4 h-5 k)$
$\therefore \quad$ Locus of $(h, k)$ is $20\left(x^{2}+y^{2}\right)-36 x+45 y=0$
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