Question

Prove that the locus of the centre of a circle, which intercepts a chord of given length $2a$ on the axis of $x$ and passes through a given point on the axis of $y$ distant $b$ from the origin, is the curve $x^2-2yb+b^2=a^2.$

Answer 1

Let the centre of the circle be $O(h,k)$ and it passes through $C(0,b)$

Length of chord $AB=2a$

$r=OCr=\sqrt{{(h-0)}^2+{(k-b)}^2}{r}^2={\left(h\right)}^2+{(k-b)}^2$

In $△OAX,$ ${OA}^2={AX}^2+{OX}^2$

$r^2=a^2+k^2{\left(h\right)}^2+{(k-b)}^2=a^2+k^2{h}^2+k^2-2bk+b^2=a^2+k^2{h}^2-2bk+b^2-a^2=0$

Replacing $h$ by $x$ and $k$ by $y$

$x^2-2by+b^2-a^2=0$

is the required locus of centre