Question

Find the equation of the circle passing through $(0,0)$ and making intercepts $a$ and $b$ on the coordinates axes.

Answer 1

Let the equation of circle be $(x-h{)}^2+(y-k{)}^2=r^2$ ____________ (1)

As center is passing through $(0,0)$

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

$\Rightarrow h^2+k^2=r^2$_______ (2)

Also $a$ & $b$ are intercept on coordinates axes

$\therefore$ Circle also passes through $(a,0)$ & $(0,b)$

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

But $r^2=h^2+k^2$

$\therefore a^2-2ah+h^2+k^2=h^2+k^2$

$a=2h\to h=\frac{a}{2}$

$\therefore$ Also, $(0-h{)}^2+(b-k{)}^2=h^2+k^2$

$h^2+b^2+k^2-2bk-h^2+k^2$

$b=2k\Rightarrow k=b/2$___________ (3)

Substituting (2) in (1) We get

$(x-h{)}^2+(y-k{)}^2=h^2+k^2$ ________________ (4)

Substituting (3) in (4) we get

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

$x^2-ax+\frac{a^2}{4}+y^2-by+\frac{b^2}{4}=\frac{a^2}{4}+\frac{b^2}{4}$

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