Once you get circumference $ = 20\pi$, just divide by $4$ to get the quarter circumference.

Answer is $20\pi / 4 = 5\pi$. simple.

However, the title of the question seems to be misleading. The question in the body seems to have nothing to do with the area and the circumference (perimeter?) of the quadrant.

Circle $A$, with radius $r$, gets back to its starting point when $A$'s centre completes one rotation (around the centre of circle $B$ with radius $R$). Clearly $A$'s centre traverses a circular path of radius = $R+r$.

Now, Physics to the rescue.

Let the speed of $A$'s centre = $v$

Let the angular speed of $A$'s rotation around its own centre = $\omega$

Since, **$A$ ***rolls* over $B$ $\implies$ $v = \omega r$ (Assuming $B$ is fixed, the condition of rolling is that the point of contact is at rest).

Let the time taken by $A$'s centre to complete one rotation be $t$.

Then, $2\pi (R+r) = vt$

$\implies t = \frac{2\pi (R+r)}{v}$

Total angular distance traversed by $A$ around its centre in the same duration: $\theta = \omega t$

Using the above results, we get $\theta = \frac{v}{r} \frac{2\pi (R+r)}{v} = \frac{2\pi (R+r)}{r}$

In this time $t$, $A$ completes, say, $N$ rotations around its centre.

$\implies N = \frac{\theta}{2\pi} = \frac{(R+r)}{r}$

Now, In your case $ r = \frac{R}{3}$

Using this information, **$ N = 4$**

## Best Answer

Circle $C_1: x^2+y^2+4x+22y+l=0$ has its center $(-2, -11)$ & a radius $\sqrt{(-2)^2+(-11)^2-l}=\sqrt{125-l}$

Similarly, circle $C_2: x^2+y^2-2x+8y-m=0$ has its center $(1, -4)$ & a radius $\sqrt{(1)^2+(-4)^2-(-m)}=\sqrt{m+17}$

Now, solving the equations of circles $C_1$ & $C_2$ by substituting the value of $(x^2+y^2)$ from $C_2$ into $C_1$, we get the $\color{blue}{\text{equation of common chord}}$ as follows $$(2x-8y+m)+4x+22y+l=0$$ $$\color{blue}{6x+14y+(l+m)=0}\tag 1$$

Now, since the circumference of circle $C_2$ is bisected by the circle $C_1$ hence the center $(1, -4)$ of circle $C_2$ must lie on the common chord or in other words, the common chord must pass through the center of circle $C_2$

Now, satisfying the above equation of common chord by center point $(1, -4)$ as follows $$6(1)+14(-4)+(l+m)=0$$ $$6-56+l+m=0$$ $$\bbox[5px, border:2px solid #C0A000]{l+m=50}$$