Question

If $y=mx$ be the equation of a chord of a circle whose radius is $a$, the origin of coordinates being one extremity of the chord and the axis of $x$ being a diameter of the circle, prove that the equations of a circle of which this chord is the diameter is $(1+m^2)(x^2+y^2)-2a(x+my)=0$.

Answer 1

Equation of chord is $y=mx$

Let the end of chord be $(h,mh)$, other end is $(0,0)$.

Equation of circle is $(x-a{)}^2+y^2=a^2$, and $(h,mh)$ lies on the circle

$\therefore (h-a{)}^2+m^2{h}^2=a^2$

$\Rightarrow h^2-2ah+m^2{h}^2=0$

$\Rightarrow h-2a+m^{2}h=0$

$\Rightarrow h=\frac{2a}{1+m^2}$

Equation of circle with $(0,0)$ and $(h,mh)$ diameter will be

$(x-h)x+(y-mh)y=0$

$\Rightarrow x^2-hx+y^2-mhy=0$

$\Rightarrow x^2+y^2=h(x+my)$

$(1+m^2)(x^2+y^2)=2a(x+my)$