Question

Find the equation of a circle of radius $5$ whose centre lies on $x-axis$ and passes through the point $(2,3)$.

Answer 1

Since the centre lies on $x$-axis, let the coordinates of centre be $(h,0)$.

Also, radius of circle $=5$

Equation of circle-

${(x-h)}^2+{(y-0)}^2=5^2$

${(x-h)}^2+y^2=25$

Also the circle passes through the point $(2,3)$.

Therefore,

${(2-h)}^2+3^2=25$

$\Rightarrow 4+h^2-4h=25-9$

$h^2-4h-12=0$

$h^2-6h+2h-12=0$

$(h-6)(h+2)=0$

$\Rightarrow h=6$ or $h=-2$

Case I:- $h=6$

Equation of circle-

${(x-6)}^2+y^2=25$

$x^2-12x+y^2+11=0$

Case II:- $h=-2$

Equation of circle-

${(x+2)}^2+y^2=25$

$x^2+4x+y^2-21=0$