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Question: Answered & Verified by Expert
The locus of the mid-point of the chord if 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
MathematicsCircleBITSATBITSAT 2022
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:
2363 Upvotes Verified Answer
The correct answer is:
$20\left(x^2+y^2\right)-36 x+45 y=0$
Let $P\left(t, \frac{4 t-20}{5}\right)$ be a point on the line
$4 x-5 y=20$. Then the chord of contact of tangents drawn from $P$ to the circle $x^2+y^2=9$ is
$t x+\left(\frac{4 t-20}{5}\right) y=9\quad \quad \dots(i)$
Let $\theta(h, k)$ be the mid-point of this chord of contact, then its equation is also $h x+k y=h^2+k^2\quad \quad \dots(ii)$
$\left[\right.$ using $\left.T=S^{\prime}\right]$
Clearly Eqs. (i) and (ii) represent the same line
$\begin{aligned}
& \therefore \frac{t}{h}=\frac{4 t-20}{5 k}=\frac{9}{h^2+k^2} \\
& \Rightarrow \frac{t}{h}=\frac{9}{h^2+k^2} \text { and } \frac{t}{h}=\frac{4 t-20}{5 k} \\
& \Rightarrow t=\frac{9 h}{h^2+k^2} \text { and } t=\frac{20 h}{4 h-5 k} \\
& \Rightarrow \frac{9 h}{h^2+k^2}=\frac{20 h}{4 h-5 k} \\
& \Rightarrow h\left\{20\left(h^2+k^2\right)-36 h+45 k\right\}=0 \\
& x=0, \text { or }\left[20\left(x^2+y^2\right)-36 x+45 y\right]=0
\end{aligned}$

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