Question

The radius of the circle $r^2-2r(3cos\theta +4sin\theta )=75$ is

• A. $5$
• B. $5\sqrt{2}$
• C. $10$
• D. $10\sqrt{2}$

Answer 1

The correct option is C $10$

put $\cos \theta =\frac{x}{r}$ & $\sin \theta =\frac{y}{r}$ in $r^2-2r(3cos\theta +4sin\theta )-75=0$
where, $r=\sqrt{x^2+y^2}$
$x^2+y^2-2r(\frac{3x}{r}+\frac{4y}{r})-75=0$
$\Rightarrow x^2+y^2-6x-8y-75=0$
Therefore, radius $=\sqrt{3^2+4^2+75}=10$

Thus the radius of the circle is $10$