Question

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

Answer 1

Given a circle of radius $r=5$
Given center lies on $x$-axis.

So, let the center be $(h,0)$
Equation of circle is
$(x-h{)}^2+y^2=25$ ....(1)
Since, it passes through $(2,3)$
$\Rightarrow (2-h{)}^2+3^2=25$
$\Rightarrow (2-h{)}^2=16$
$\Rightarrow 2-h=\pm 4$
$\Rightarrow h=-2,h=6$
When $h=-2$, the equation of circle is
$(x+2{)}^2+y^2=25$
$\Rightarrow x^2+4x+4+y^2=25$
$\Rightarrow x^2+y^2+4x-21=0$
When $h=6$, the equation of circle is
$(x-6{)}^2+y^2=25$
$\Rightarrow x^2-12x+36+y^2=25$
$\Rightarrow x^2+y^2-12x+11=0$