Question

Find the equation of the circle concentric with the circle $x^2+y^2-8x-12y+15=0$ and passing through the point $(5,4)$.

Answer 1

Equation of circle is

$x^2+y^2-8x-12y+15=0$

On comparing that,

$x^2+y^2+2gx+2fy+c=0$

So,

$(g,f)=(-4,-6)$

$(-g,-f)=(4,6)=(h,k)$

Now, it passes through the point $(5,4)$

So,

$Radius=\sqrt{{(4-5)}^2+{(6-4)}^2}$

$Radius=\sqrt{1+4}$

$Radius=\sqrt{5}$

So, equation of circle is,

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

$\Rightarrow {(x-4)}^2+{(y-6)}^2={\left(\sqrt{5}\right)}^2$

$\Rightarrow x^2+16-8x+y^2+36-12y=25$

$\Rightarrow x^2+y^2-8x-12y+52=25$

$\Rightarrow x^2+y^2-8x-12y+27=0$

Hence, this is the answer.