Question

The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line $4x-5y=20$ to the circle $x^2+y^2=9$ is

• A. $20(x^2+y^2)-36x+45y=0$
• B. $20(x^2+y^2)+36x-45y=0$
• C. $20(x^2+y^2)-20x+45y=0$
• D. $20(x^2+y^2)+20x-45y=0$

Answer 1

Answer: (A)

$20(x^2+y^2)-36x+45y=0$

Given: $4x-5y=20$

Let midpoint of the chord of contact be $(h,k).$

Then Equation of the chord is with the above midpoint will be given as $hx+ky=h^2+k^2.....\left(1\right)$ (using the general equation of midpoint chord relation)

Let any point on the line $4x-5y=20$ be P $(\alpha ,\frac{4}{5}\alpha -4)$

Then using the relation of the equation of chord of contact from the given outside point P we get,

$\alpha x+(\frac{4}{5}\alpha -4)y=9.....\left(2\right)$

Since (1) and (2) represent the same equation of chord, equating the coefficients

$\frac{h}{\alpha }=\frac{k}{\frac{4}{5}\alpha -4}=\frac{h^2+k^2}{9}$

$\Rightarrow \alpha =\frac{9h}{h^2+k^2}.....\left(3\right)$

Also $\frac{4}{5}\alpha -4=\frac{9k}{h^2+k^2}$

$\Rightarrow \frac{4}{5}\alpha =\frac{9k+4(h^2+k^2)}{h^2+k^2}$

$\Rightarrow \alpha =\frac{45k+20(h^2+k^2)}{4(h^2+k^2)}.....\left(4\right)$

Eliminate ${}^{\prime }{\alpha }^{\prime }$ from '3' and '4' to get the locus.

$4\times 9h=45k+20(h^2+k^2)$

$\Rightarrow 20(h^2+k^2)-36h+45k=0$

$\therefore$ Locus of the midpoint is $20(x^2+y^2)-36x+45y=0$