The locus of the midpoints of the chord of the circle,

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 :

  1. $\left(x^{2}+y^{2}\right)^{2}-16 x^{2}+9 y^{2}=0$

  2. $\left(x^{2}+y^{2}\right)^{2}-9 x^{2}+144 y^{2}=0$

  3. $\left(x^{2}+y^{2}\right)^{2}-9 x^{2}-16 y^{2}=0$

  4. $\left(x^{2}+y^{2}\right)^{2}-9 x^{2}+16 y^{2}=0$


Correct Option: , 4

Solution:

$y-k=-\frac{h}{k}(x-h)$

$\mathrm{ky}-\mathrm{k}^{2}=-\mathrm{hx}+\mathrm{h}^{2}$

$\mathrm{hx}+\mathrm{ky}=\mathrm{h}^{2}+\mathrm{k}^{2}$

$\mathrm{y}=-\frac{\mathrm{hx}}{\mathrm{k}}+\frac{\mathrm{h}^{2}+\mathrm{k}^{2}}{\mathrm{k}}$

tangent to $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$

$\mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~m}^{2}-\mathrm{b}^{2}$

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

$\left(x^{2}+y^{2}\right)^{2}=9 x^{2}-16 y^{2}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now