Question

The equation to the circle having y$=$mx as a diameter where y$=$ mx is a chord of the circle through the origin of status a and having the x-axis diameter is ?

Answer 1

We have a circle which has a chord $y=mx$ of radius ${}^{\prime }{a}^{\prime }$ and the diameter is along $x-$axis, also one end of the chord is origin. The figure for the above scenario will be

$\therefore$ equation of circle will be

${(x-a)}^2+y^2=a^2$

$\Rightarrow x^2+y^2-2ax=0$

Now let's find the coordinates of point ${}^{\prime }{A}^{\prime }$, for which we have

$\therefore$ to solve $y=mx$ and $x^2+y^2-2ax=0$ simultaneously

$\therefore$ ${\left(mx\right)}^2+x^2-2ax=0$

$\Rightarrow m^2{x}^2-2ax=0$

$\Rightarrow x^2(1+m^2)-2ax=0$

$\Rightarrow x[x(1+m^2)-2a]=0$

$\therefore$ $x=0$ or $x=\frac{2a}{1+m^2}$

$\therefore$ $y=0$ or $y=\frac{2ma}{1+m^2}$

$\therefore$ $O(0,0)$ and $A(\frac{2a}{1+m^2},\frac{2am}{1+m^2})$

Now with $OA$ as diameter for another circle

$\therefore$ the centre of the circle $=(\frac{a}{1+m^2},\frac{2am}{1+m^2})$

radius $=\frac{1}{2}\sqrt{{\left(\frac{2a}{1+m^2}\right)}^2+{\left(\frac{2am}{1+m^2}\right)}^2}=\frac{1}{2}\sqrt{{\left(\frac{2a}{1+m^2}\right)}^2(1+m^2)}$

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

$\therefore$ equation of circle : ${(x-\frac{a}{1+m^2})}^2+(y-\frac{am}{1+m^2})=\frac{a^2}{1+m^2}$

Simplifying the equation we obtain

$(1+m^2)(x^2+y^2)-2ax-2amy=0$