Q98 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Mathematics

The locus of the mid-points of the chords of the circle $x^2+y^2=16$ which are the tangents to the hyperbola $9 x^2-16 y^2=144$ is

  • A
    $3 x^2-4 y^2=16\left(x^2+y^2\right)$
  • B
    $4 x^2-3 y^2=9\left(x^2+y^2\right)$
  • Correct answer
    C
    $16 x^2-9 y^2=\left(x^2+y^2\right)^2$
  • D
    $16 x^2-9 y^2=4\left(x^2+y^2\right)$

Solution

Equation of chord with mid-point $(h, k)$ to the circle $x^2+y^2=16$ is $T=S_1$ $$ \begin{gathered} h x+k y-16=h^2+k^2-16 \\ y=-\frac{h}{k} x+\left(\frac{h^2+k^2}{k}\right) \end{gathered} $$ For hyperbola $9 x^2-16 y^2=144$ $$ \Rightarrow \quad \frac{x^2}{16}-\frac{y^2}{9}=1 $$ As we know that, condition of tangency of a line $y=m x+c$ is $a^2 m^2-b^2=c^2$ So, $$ \begin{aligned} \text { So, } & & 16\left(-\frac{h}{k}\right)^2-9 & =\left(\frac{h^2+k^2}{k}\right)^2 \\ \Rightarrow & & 16 h^2-9 k^2 & =\left(h^2+k^2\right)^2 \end{aligned} $$ Replace $h, k$ by $x, y$ $$ \therefore \quad 16 x^2-9 y^2=\left(x^2+y^2\right)^2 $$

Hence, the correct answer is option: 'C'.

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