Question

Locus of the mid point of the chord of circle $x^2+y^2=16$ which is subtending right angle at the point $(5,0)$ is

• A. $x^2+y^2+4x+\frac{5}{2}=0$
• B. $x^2+y^2-5x+\frac{9}{2}=0$
• C. $x^2+y^2+5x+7=0$
• D. $x^2+y^2+2x+5=0$

Answer 1

The correct option is B $x^2+y^2-5x+\frac{9}{2}=0$

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

The equation of chord is $hx+ky=h^2+k^2$

The end points of chord be $(a,b),(p,q)$

By using the condition of perpendicularity , we get $\left(\frac{b-0}{a-5}\right)\times \left(\frac{q-0}{p-5}\right)=-1$

$\Rightarrow ap+bq+25=5(a+p)$

Then using the equation of chord and separately eliminating $x,y$ to obtain quadratics

$\lambda x^2-2\lambda hx+{\lambda }^2+16k^2=0$ and $\lambda y^2-2\lambda ky+{\lambda }^2-16h^2=0$ , where $\lambda =h^2+k^2$

Using the values of product of roots and sum of roots , the locus is

$x^2+y^2-5x+\frac{9}{2}=0$

Therefore correct option is $B$