Q.Two circles, each of radius 7 cm, intersect each other. The distance between their centres is $7\sqrt{2}$ cm. Find the area of the portion common to both the circles.

Step 1: Let $O_1,O_2$ be the centres, $r=7$ cm, $d=O_1O_2=7\sqrt2$ cm. The chord of intersection is perpendicular to $O_1O_2$ at its midpoint $M$, with $O_1M=\frac{d}{2}=\frac{7\sqrt2}{2}$.

Step 2: In right triangle $O_1MP$ (P an intersection point), $\cos\alpha=\frac{O_1M}{O_1P}=\frac{7\sqrt2/2}{7}=\frac{\sqrt2}{2}\Rightarrow \alpha=45°$. The full central angle subtended at each centre by the chord is $2\alpha=90°=\frac{\pi}{2}$.

Step 3: Area of one circular segment (the overlap contributed by one circle) $=$ sector area $-$ triangle area $=\frac12 r^2(2\alpha) - \frac12 r^2\sin(2\alpha) = \frac12(49)\frac{\pi}{2}-\frac12(49)\sin90° = \frac{49\pi}{4}-\frac{49}{2}$.

Step 4: Total common area $= 2\times$ segment area $= 2\left(\frac{49\pi}{4}-\frac{49}{2}\right) = \frac{49\pi}{2}-49 = \frac{49(\pi-2)}{2}$ sq.cm $\approx 27.97$ sq.cm.

Find the central angle subtended by the common chord at each circle's centre using the triangle formed by the two radii and half the inter-centre distance, then double the resulting circular-segment area.

• Forgetting to DOUBLE the single-circle segment area, since the overlap is made of two equal segments (one from each circle).
• Sign/algebra slip finding cos α = (d/2)/r and then the angle α.
• Confusing the segment formula ½r²(θ−sinθ) with the sector formula ½r²θ alone.