Question

Determined the equation of the circles passes through the point $(a,0)$ and $(0,a)$ and whose radius is as small as possible . Show that it will pass through the origin

Answer 1

The equation of the circle passing through two points by S + $\lambda$P = 0 is

$x(x-a)+y(y-a)+\lambda (x+y-a)=0$

or $x^2+y^2-(a-\lambda )-x(a-\lambda )y-\lambda a=0$

We have to choose $\lambda$ such that its radius is least

$r^2={\left(\frac{a-\lambda }{2}\right)}^2+{\left(\frac{a-\lambda }{2}\right)}^2+\lambda a=0$

$\frac{1}{2}(a^2+{\lambda }^2-2a\lambda +2a\lambda )=\frac{1}{2}(a^2+{\lambda }^2)$

It will be least when $\lambda$ = 0 and hence the required circle is $x^2+y^2-ax-by=0$ which passes through origin.