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

Let the equation of circle is $$ x^{2}+y^{2}+2 g x+2 f y+c=0 $$ When, circle (i) passes through the onigin Then, $\quad c=0$ When, circle (i) passes through the point (2,6). Then $4+36+4 g+12 f+0=0$ $\Rightarrow \quad 4 g+12 f+40=0$ $\Rightarrow g+3 f=-10$ When, circle (1) passes through the point (6,2) Then, $36+4+12 g+4 t+0=0$ [from Eq. $\Rightarrow \quad 12 g+4 f+40=0$ $\Rightarrow \quad 3 g+t+10=0$ On solving Eqs. (iii) and (iv), we get $g=\frac{-5}{2}$ and $f=\frac{-5}{2}$ $\therefore$ Equation of circle becomes $$ x^{2}+y^{2}-5 x-5 y=0 $$ Circle cut the $x$ -axis. So, put $y=0$ in Eq. $(v)$, we get $$ x^{2}-5 x=0 \Rightarrow x(x-5)=0 $$ $\Rightarrow \quad x=5$ So, the circle cut the $x$ -axis at point $P(5,0)$ $\therefore$ The length of $O P=5$