Question

The locus of the midpoints of the chord of the circle, $x^2+y^2=25$ which is tangent to the hyperbola, $\frac{x^2}{9}-\frac{y^2}{16}=1$is

• A. ${(x^2+y^2)}^2–16x^2+9y^2=0$
• B. ${(x^2+y^2)}^2–9x^2+144y^2=0$
• C. ${(x^2+y^2)}^2–9x^2–16y^2=0$
• D. ${(x^2+y^2)}^2–9x^2+16y^2=0$

Answer 1

The correct option is D

${(x^2+y^2)}^2–9x^2+16y^2=0$

Explanation of the correct option.

Compute the locus:

Since the tangent of hyperbola $y=mx\pm \sqrt{(9m^2–16)}$……………….$\left(1\right)$

which is a chord of a circle with mid-point $\left(h,k\right)$.

So, the equation of chord $T=S_1$

$hx+ky=h^2+k^2$

$\Rightarrow$ $y=-\left(\frac{hx}{k}\right)+\left(\frac{h^2+k^2}{k}\right)$……………….$\left(2\right)$

Compare equation $\left(1\right)$ and $\left(2\right)$,

$m=–\frac{h}{k}$ and $\sqrt{9m^2–16}=\frac{h^2+k^2}{k}$

$\Rightarrow 9\left(\frac{h^2}{k^2}\right)–16=\frac{h^2+k^2}{k^2}$

Therefore the locus is

$9x^2–16y^2={(x^2+y^2)}^2$

Hence, option $\left(\mathrm{D}\right)$ is the correct option.