Question

The centre of the circle
$r^2-4r(\cos \theta +\sin \theta )-4=0$ in cartesian coordinates is :

• A. $(1,1)$
• B. $(-1,-1)$
• C. $(2,2)$
• D. $(-2,-2)$

Answer 1

The correct option is C $(2,2)$

Put $\cos \theta =\frac{x}{r}$ & $\sin \theta =\frac{y}{r}$ in $r^2-4r(\cos \theta +\sin \theta )-4=0$
where, $r=\sqrt{x^2+y^2}$
$x^2+y^2-4r(\frac{x}{r}+\frac{y}{r})-4=0$
$\Rightarrow x^2+y^2-4x-4y-4=0$
Therefore, center of circle is$(2,2)$.
Ans: C