Find the equation of the circle passing through (0, 0)

Solution:

Let the equation of the required circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.

Since the circle passes through $(0,0)$,

$(0-h)^{2}+(0-k)^{2}=r^{2}$

$\Rightarrow h^{2}+k^{2}=r^{2}$

The equation of the circle now becomes $(x-h)^{2}+(y-k)^{2}=h^{2}+k^{2}$.

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,

$(a-h)^{2}+(0-k)^{2}=h^{2}+k^{2}$ $. .(1)$

$(0-h)^{2}+(b-k)^{2}=h^{2}+k^{2}$    (2)

From equation (1), we obtain

$a^{2}-2 a h+h^{2}+k^{2}=h^{2}+k^{2}$

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

$\Rightarrow a(a-2 h)=0$

$\Rightarrow a=0$ or $(a-2 h)=0$

However, $a \neq 0$; hence, $(a-2 h)=0 \Rightarrow h=\frac{a}{2}$.

From equation (2), we obtain

$h^{2}+b^{2}-2 b k+k^{2}=h^{2}+k^{2}$

$\Rightarrow b^{2}-2 b k=0$

$\Rightarrow b(b-2 k)=0$

$\Rightarrow b=0$ or $(b-2 k)=0$

However, $b \neq 0$; hence, $(b-2 k)=0 \Rightarrow k=\frac{b}{2}$

Thus, the equation of the required circle is

$\left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\left(\frac{a}{2}\right)^{2}+\left(\frac{b}{2}\right)^{2}$

$\Rightarrow\left(\frac{2 x-a}{2}\right)^{2}+\left(\frac{2 y-b}{2}\right)^{2}=\frac{a^{2}+b^{2}}{4}$

$\Rightarrow 4 x^{2}-4 a x+a^{2}+4 y^{2}-4 b y+b^{2}=a^{2}+b^{2}$

$\Rightarrow 4 x^{2}+4 y^{2}-4 a x-4 b y=0$

$\Rightarrow x^{2}+y^{2}-a x-b y=0$