A line $l$ meets the circle $x^2+y^2=61$ in $A, B$ and $P(-5,6)$ is such that $P A=P B=10$. Then, the equation of $l$ is

Question

Aicharya · Accepted Answer

Since, the line $l$ meets the circle $x^2+y^2=61$ at two points $A$ and $B$.

$ \because P A=P B=10 $ $\therefore P M$ is perpendicular to $A B$ and $O M$ is perpendicular to the chord $A B$, therefore line $O P$ is perpendicular to $A B$. $\therefore$ The slope of line $P O$ is $-\frac{6}{5}$. Now, taking option (c). Let the required line be $5 x-6 y+11=0$ its slope is $\frac{6}{5}$. Now, $ \frac{5}{6} \times\left(-\frac{5}{6}\right)=-1 $ Therefore our assumption is true

A line $l$ meets the circle $x^2+y^2=61$ in $A, B$ and $P(-5,6)$ is such that $P A=P B=10$. Then, the…

Solution