Question

The area (in $\text{sq.units}$) enclosed by the curve $|x^2-4x+y^2-6y|<12$ is equal to

• A. $6\pi$
• B. $12\pi$
• C. $24\pi$
• D. $50\pi$

Answer 1

The correct option is C $24\pi$

Given curve:
$|x^2-4x+y^2-6y|<12\Rightarrow |(x-2{)}^2+(y-3{)}^2-13|<12\Rightarrow -12<(x-2{)}^2+(y-3{)}^2-13<12\Rightarrow 1<(x-2{)}^2+(y-3{)}^2<25$
Which shows the region between two co-axial ring, of radius $1,5\text{units}.$

So, Required area $=\pi (5^2-1^2)=24\pi \text{sq.units}$