Question

Find the number of circle having radius $5$ and passing through the points $(-2,0)$ and $(4,0)$

Answer 1

let center of circle is $(h,k)$

so ${(h+2)}^2+k^2=25$ .......$\left(1\right)$

and ${(h-4)}^2+k^2=25$ .......$\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$

${(h+2)}^2+k^2={(h-4)}^2+k^2$

$\Rightarrow {(h+2)}^2={(h-4)}^2$

$\Rightarrow {(h+2)}^2-{(h-4)}^2=0$ is of the form $a^2-b^2$

$\Rightarrow (h+2+h-4)(h+2-h+4)=0$ where $a^2-b^2=(a+b)(a-b)$

$\Rightarrow 6(2h-2)=0$ or $h=1$

Substituting $h=1$ in eqn$\left(1\right)$ we get

${(h+2)}^2+k^2=25$

$\Rightarrow 3^2+k^2=25$

$\Rightarrow k^2=25-9=16$

$\Rightarrow k=\pm 4$

So two circles are possible.