Find the equation of the circle with passing through three points $(0,2),(3,0) and (3,2)$

Concept: If a circle passes through the three points, then they must satisfy the equation of the circle.

Here, three points on the circle are $\text{A}\equiv (0,2),\text{B}\equiv (3,0),\text{C}\equiv (3,2)$

Let $\text{P}\equiv (\text{x},\text{y})$   be its centre, then $\left|\text{PA}\right|=\left|\text{PB}\right|=\left|\text{PC}\right|$

$\Rightarrow \sqrt{{(x-0)}^2+{(y-2)}^2}=\sqrt{{(x-3)}^2+{(y-0)}^2}=\sqrt{{(x-3)}^2+{(y-2)}^2}$

$\Rightarrow \sqrt{x^2+y^2+4-4y}=\sqrt{x^2+9-6x+y^2}=\sqrt{x^2+9-6x+y^2+4-4y}$

$\Rightarrow 4-4y=9-6x=9-6x+4-4y$

Taking first and third parts of the above equation,

$4-4y=9+4-6x-4y\Rightarrow 6x=9\Rightarrow x=\frac{9}{6}=\frac{3}{2}$

When taking second and third parts of the above equation, we get

$9-6x=9+4-6x-4y\Rightarrow 4y=4\Rightarrow y=1\therefore P\equiv \left(\frac{3}{2},1\right)$

Thus, radius 

$\text{r}=\text{PA}=\sqrt{{\left(\frac{3}{2}\right)}^2+{(1-2)}^2}=\sqrt{\frac{9}{4}+1}=\frac{\sqrt{5}}{2}$Now, equation of the circle will be 

${\left(x-\frac{3}{2}\right)}^2+{(y-1)}^2={\left(\frac{\sqrt{5}}{2}\right)}^2$

$\begin{array}{l}\Rightarrow x^2+y^2-3x-2y+1-\frac{5}{4}+\frac{9}{4}=0\\ \Rightarrow x^2+y^2-3x-2y=0\end{array}$

Hence, equation of the circle is

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