Question

The radius of the circle
$r^2-2\sqrt{2}r(\cos \theta +\sin \theta )-5=0$ is

• A. 9
• B. 5
• C. 3
• D. 2

Answer 1

The correct option is C 3

Put $\cos \theta =\frac{x}{r}$ & $\sin \theta =\frac{y}{r}$ in $r^2-2\sqrt{2}r(3\cos \theta +4\sin \theta )-5=0$
where, $r=\sqrt{x^2+y^2}$
$x^2+y^2-2\sqrt{2}r(\frac{x}{r}+\frac{y}{r})-5=0$
$\Rightarrow x^2+y^2-2\sqrt{2}x-2\sqrt{2}y-5=0$
Therefore, radius $=\sqrt{{\sqrt{2}}^2+{\sqrt{2}}^2+5}=3$
Ans: C