Question

find the equation of the circle passing through (1,0) and (0,1) and having the smallest possible radius

Answer 1

Let the equation of circle is :

$x^2+y^2+2gx+2fy+c=0$

This passes through (1,0) and (0,1), therefore,

$1+2g+c=0$

$1+2f+c=0$

$⟹g=f$ and $c=-2g-1$

Radius r of circle is given by :

$r^2=g^2+f^2-c$

$=2g^2+2g+1=S$

$\frac{dS}{dg}=4g+2$

$\frac{d^{2}S}{d{g}^2}=4>0$

Now, $\frac{dS}{dg}=0$

$g=-\frac{1}{2}$

$c=0$

Hence, equation of required circle is $x^2+y^2-x-y=0$.