Question

Find the equations of the circles passing through two points on y-axis at distance $3$ from the origin and having radius $5$.

Answer 1

According to the given information, circle is passing through two point $(0,3),(0,-3)$ and radius is $5$

Equation of circle is,
$(x-h{)}^2+(y-k{)}^2=r^2$
$(x-h{)}^2+(y-k{)}^2=25\dots \left(i\right)$
It is passing through $(0,3)$
$\Rightarrow (0-h{)}^2+(3-k{)}^2=25$
$\Rightarrow h^2+(3-k{)}^2=25\dots (2)$

It is also passing through $(0,-3)$
$(0-h{)}^2+(-3-k{)}^2=25$
$h^2+(3+k{)}^2=25\dots (3)$
From equation(2) - (3),
$(3-k{)}^2-(3+k{)}^2=0$
$\Rightarrow (3-k+3+k)(3-k-3-k)=0$ [Since, $a^2-b^2)=(a-b\left)\right(a+b)$]
$\Rightarrow 6(-2k)=0$
$\therefore k=0$
Put $k=0$ in equation (2)
$h^2+(3-0{)}^2=25$
$\Rightarrow h^2=25-9$
$\Rightarrow h^2=16$
$\therefore h=\pm 4$
Put $h=4$ and $k=0$ in equation (1),
$(x-4{)}^2+y^2=25\Rightarrow x^2-8x+y^2=9$
Put $h=-4$ and $k=0$ in equation (1),
$(x+4{)}^2+y^2=25\Rightarrow x^2+8x+y^2=9$
$\therefore x^2\pm 8x+y^2=9$