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Question: Answered & Verified by Expert
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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Options:
  • A $20\left(x^{2}+y^{2}\right)-36 x+45 y=0$
  • B $20\left(x^{2}+y^{2}\right)+36 x-45 y=0$
  • C $36\left(x^{2}+y^{2}\right)-20 x+45 y=0$
  • D $36\left(x^{2}+y^{2}\right)+20 x-45 y=0$
Solution:
2294 Upvotes Verified Answer
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 $20\left(x^{2}+y^{2}\right)-36 x+45 y=0$

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