CAT 2022 Slot 3 · Quant · Question 18

A circle on BC as diameter passes through A exactly when the angle at A is a right angle.

So triangle ABC is right-angled and isosceles, with the two equal sides of 8 cm meeting at A.

By Pythagoras, B{C}^{2}={8}^{2}+{8}^{2}=128, so BC=8\sqrt{2}.

The first circle therefore has radius \frac{BC}{2}=4\sqrt{2}, centred at the midpoint M of BC.

The second circle is centred at A with radius AB=8, so it also passes through B and C.

Both circles pass through B and C, so those two points are exactly where they cross.

The overlap is the lens between them, split by the chord BC into two pieces.

On the A side of BC, the piece is half of the small circle, because BC is that circle's diameter.

That half has area \frac{1}{2}\pi{\left( 4\sqrt{2}\right)}^{2}=\frac{1}{2}\pi \times 32=16\pi.

On the far side of BC, the piece is the segment of the big circle cut off by the chord BC.

The angle BAC is {90}^{∘}, so that segment is a quarter circle minus triangle ABC.

Its area is \frac{1}{4}\pi \times{8}^{2}-\frac{1}{2}\times 8\times 8=16\pi -32.

Adding the two pieces, the overlap is 16\pi +\left( 16\pi -32 \right)=32\pi -32=32\left( \pi -1 \right) — the second option.

Shortcut. Each piece is a standard shape once BC is recognised as a diameter of one circle and a right-angle chord of the other. Note also that the overlap must exceed the small circle's half, 16\pi, which already rules out the two smaller options.