A circle passes through the points (2,3) and (4,5).

Question

A circle passes through the points $(2,3)$ and $(4,5)$. If its centre lies on the line $y-4x+3=0$, then its radius is equal to

Accepted Answer

Centre $(\alpha, \beta)$ lies on line $y-4 x+3=0$

Then, $\quad \beta=4 \alpha-3$

And

$$
\text { Radius }=\sqrt{(\alpha-2)^2+(\beta-3)^2}=\sqrt{(\alpha-4)^2+(\beta-5)^2}
$$

$$
\begin{aligned}
& \alpha^2+\beta^2+ 13-4 \alpha-6 \beta=\alpha^2+\beta^2+41-8 \alpha-10 \beta \\
& 4 \alpha+4 \beta=28 \Rightarrow \alpha+\beta=7 \\
& \Rightarrow \alpha+4 \alpha-3=7 \\
& \Rightarrow \alpha=2, \beta=5
\end{aligned}
$$

Therefore, $\quad$ Radius $=\sqrt{(2-2)^2+(5-3)^2}=2$

Solutions · 1