Question

A circle passing through $(0,0)$, $(2,6)$, $(6,2)$ cut the $x$-axis at the point $P\ne (0,0)$. Then, the length of $OP$, where $O$ is the origin, is

• A. $\frac{5}{2}$
• B. $\frac{5}{\sqrt{2}}$
• C. $5$
• D. $10$

Answer 1

The correct option is C $5$

Let the equation of circle is
$x^2+y^2+2gx+2fy+c=\mathrm{0.....}\left(i\right)$

When circle $\left(i\right)$ passes through the origin.
Then, $c=0$

When, circle $\left(i\right)$ passes through the point $(2,6)$.

Then, $4+36+4g+12f+0=0$
$\Rightarrow$ $4g+12f+40=0$
$\Rightarrow$ $g+3f=-\mathrm{10.....}\left(iii\right)$

When circle $\left(i\right)$ passes through the point $(6,2)$

Then, $35+4+12g+4f+0=0$ (from eq $\left(i\right)$)
$\Rightarrow$ $12g+4f+40=0$
$\Rightarrow$ $3g+f+10=\mathrm{0.....}\left)\right(iv)$

On solving eqs. $\left(iii\right)$ and $\left(iv\right)$ we get
$g=\frac{-5}{2}$ and $f=\frac{-5}{2}$
$\therefore$ Equation of circle becomes $x^2+y^2-5x-5y=\mathrm{0.......}\left(v\right)$

Circle cut the $x$-axis
So, put $y=0$ in eq $\left(v\right)$, we get
$x^2-5x=0$ $\Rightarrow$ $x(x-5)=0$
$\Rightarrow$ $x=5$
So, the circle cut the $x$-axis at point $P(5,0)$
$\therefore$ The length of $OP=5$