Question

Find the equation of that chord of the circle $x^2+y^2=15$ Which is bisected at the point (3,2).

Answer 1

Equation of the circle is given as

$x^2+y^2=15$ …………..$\left(1\right)$

$\therefore$ Centre of the circle, $C=(0,0)$ and radius, $r=\sqrt{15}$

$\therefore g=0,f=0$

Here, it is given that the chord is bisected at the point $(3,2)$

$\therefore$ Midpoint of the chord, $M=(3,2)$

The equation of the chord of the circle $\left(1\right)$ with $M(3,2)$ as midpoint is given by

$3x+2y+0(x+3)+0(y+2)=(3{)}^2+(2{)}^2$

$\Rightarrow 3x+2y=9+4$

$\Rightarrow 3x+2y=13$.