Question

A line is drawn from $O$ and intersecting the circle $x^2+y^2-6x-8y+16=0$ at $P_1$ and $Q_1$ and making angle $\alpha$ with positive $x-$axis, then ${OP}_1.{OQ}_1$ is equal to

• A. $0$
• B. $9$
• C. $16$
• D. $25$

Answer 1

The correct option is C $16$

As the center of the circle C is $(3,4)$
radius $a=3$
And if OC is making an angle $\beta$ with O$\alpha$ and $OC=\gamma$
So from equation$⟶\left(1\right)r^2-2r{r}_1\cos (\theta -{\theta }_1)+r^2-a^2=0r^2-2r\gamma \cos (\alpha -\beta )+{\gamma }^2-{\alpha }^2=0$
$\gamma \&r$ make two radii.
$\therefore O{P}_1.O{Q}_1$ $=r_1\times r_2{\gamma }^2-{\alpha }^2=25-9=16$