JEE Main 2021 (Online) 27th July Evening ShiftQuestion 1
Consider a circle $C$ which touches the $y$-axis at $(0,6)$ and cuts off an intercept $6\sqrt5$ on the $x$-axis. Then the radius of the circle $C$ is equal to :
Circle
Solution
Touches $y$-axis at $(0,6)$: centre $(r,6)$ (radius $r$). $x$-intercept $6\sqrt5$: $\left(3\sqrt5\right)^2=r^2-6^2\Rightarrow r^2=45+36=81$. $r=9$. Option (C).