If a circle passes through points $(4,0)$ and $(0,2)$ and its centre lies on $\mathrm{Y}$-axis. If the…
If a circle passes through points $(4,0)$ and $(0,2)$ and its centre lies on $\mathrm{Y}$-axis. If the radius of the circle is $r$, then the value of $r^2-r+1$ is
$25$
$21$
$20$
$10$
Solution
Let $(0, y)$ be the centre of the circle.
$\begin{aligned}
& \therefore \quad \sqrt{(0-4)^2+(y-0)^2}=\sqrt{(0-0)^2+(y-2)^2} \\
& \therefore \quad 16+y^2=(y-2)^2 \\
& \therefore \quad 16+y^2=y^2-4 y+4 \\
& \begin{aligned}
\therefore & y=-3 \\
\therefore & \text { centre of the circle is }(0,-3) . \\
\therefore & \text { Radius of the circle }=\mathrm{r}=\sqrt{(0-0)^2+(-3-2)^2} \\
& =5 \text { units } \\
\therefore & \mathrm{r}^2-\mathrm{r}+1=25-5+1=21
\end{aligned}
\end{aligned}$