Question

The locus of center of a circle which passes through the origin and cut off a length of $4$ units from the line $x=3$is

• A. $y^2+6x=0$
• B. $y^2+6x=13$
• C. $y^2+6x=10$
• D. $x^2+6y=13$

Answer 1

The correct option is B

$y^2+6x=13$

Explanation for correct option

Let the circle be $C\left(-g,-f\right)$, then the equation of the line passing through origin is $x^2+y^2+2gx+2fy=0$

Distance, $d=CA=\left|-g-3\right|=g+3$

In $△ABC,$

Using distance formula, for points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_2\right)}^2}$

$$\begin{gathered} {\left(BC\right)}^2=O{C}^2={\left(\sqrt{{\left(0-g\right)}^2+{\left(0-f\right)}^2}\right)}^2=f^2+g^2 \\ {\left(AC\right)}^2={\left(\sqrt{{\left(g+3\right)}^2+{\left(f-f\right)}^2}\right)}^2={(g+3)}^2 \\ {\left(BA\right)}^2={\left(\sqrt{{\left(g+3-(g+3)\right)}^2+{(f+2-f)}^2}\right)}^2=2^2 \end{gathered}$$

${\left(BC\right)}^2={\left(AC\right)}^2+{\left(AB\right)}^2$ [ Pythagoras theorem ]

$$\begin{gathered} f^2+g^2={(g+3)}^2+4 \\ {f}^2+g^2=g^2+9+6g+4 \\ {f}^2=6g+13 \\ {f}^2-6g-13=0 \end{gathered}$$

Eliminating $-f=y,-g=x$

$y^2+6x=13$

Hence, option B is correct.