Question

Find the equation of a circle which passes through $(4,1),(6,5)$ and having the centre on:
$4x+3y-24=0$.

Answer 1

Let $(h,k)$ be the centre of circle.

Since the centre of circle lies on the line $4x+3y-24=0$

Therefore,

$4h+3k-24=0.....\left(1\right)$

Given that the points $(4,1)$ and $(6,5)$ lies on the circle.

Therefore,

$\sqrt{{(4-h)}^2+{(1-k)}^2}=\sqrt{{(6-h)}^2+{(5-k)}^2}$

$\sqrt{16+h^2-8h+1+k^2-2k}=\sqrt{36+h^2-12h+25+k^2-10k}$

$\sqrt{h^2+k^2-8h-2k+17}=\sqrt{h^2+k^2-12h-10k+61}$

Squaring both sides, we get

$h^2+k^2-8h-2k+17=h^2+k^2-12h-10k+61$

$8h+2k-17=12h+10k-61$

$4h+8k-44=0.....\left(2\right)$

Subtracting equation $\left(1\right)$ from $\left(2\right)$, we have

$(4h+8k-44)-(4h+3k-24)=0$

$5k-20=0$

$\Rightarrow k=4$

Substituting the value of $k$ in equation $\left(2\right)$4, we have

$4h+8\times 4-44=0$

$h=\frac{12}{4}=3$

Thus the centre of circle is $(3,4)$.

Now,

Radius of circle $=\sqrt{{(4-3)}^2+{(1-4)}^2}=\sqrt{1+9}=\sqrt{10}$

Therefore,

Equation of circle is-

${(x-3)}^2+{(y-4)}^2={\left(\sqrt{10}\right)}^2$

Hence the equation of circle is ${(x-3)}^2+{(y-4)}^2=10$.