The equation of the smallest circle passing from points $(1, 1)$and $(2, 2)$and always in the first quadrant, is

 

Let the equation of the required circle be

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

It passes through $(1, 1)$ and $(2, 2).$

$\begin{array}{r}\therefore 2+2g+2f+c=0\\ 8+4g+4f+c=0\end{array}$

From (ii) and (iii), we get

$b+f=-3$and $c=4$

Let$r$be the radius of the circle. Then,

$\begin{array}{l}{r}^2=g^2+f^2-c\\ \Rightarrow r^2=g^2+f^2-4\\ \Rightarrow r^2=g^2+(3+g{)}^2-4\\ \Rightarrow r^2=2g^2+6g+5\end{array}$ $[\because g+f=-3]$

$\Rightarrow r^2=2\left\{g^2+3g+\frac{5}{2}\right\}\Rightarrow r^2=2\left\{{\left(g+\frac{3}{2}\right)}^2+\frac{1}{4}\right\}$

Clearly, $r$ is least when $g+\frac{3}{2}=0\text{ i.e. }g=-\frac{3}{2}$

Substituting the values of$g, f$and c in (i), we get

$x^2+y^2-3x-3y+4=0$