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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
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Verified Answer
The correct answer is:
$5 x-6 y+11=0$
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.
$$
\because P A=P B=10
$$
$\therefore P M$ 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.
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