Question

Find the equation to the circle which touches the axis of $y$ at the origin and passes through the point $(b,c)$.

• A. $b{x}^2+b{y}^2-(b^2+c^2)y=0$
• B. $b{x}^2+b{y}^2-(b^2+c^2)x=0$
• C. $b{x}^2+b{y}^2+(b^2+c^2)y=-1$
• D. $b{x}^2+c{y}^2-(b^2+c^2)x=1$

Answer 1

Answer: (B)

$b{x}^2+b{y}^2-(b^2+c^2)x=0$

Equation of circle which touches the $y$-axis at origin is

$x^2+y^2+2gx+d=0$

Since the circle passes through origin, $d=0$

Thus the equation becomes,

$x^2+y^2+2gx=0$ ..........(1)

The equation passes through $(b,c)$, so

$b^2+c^2+2gb=0$

Therefore, $g=\frac{-b^2-c^2}{2b}$

So, putting the value of $g$ in (1), we get

$b{x}^2+b{y}^2-(b^2+c^2)x=0$