UPSC CDS 1 2020 Math Paper

Concept:Two circles touching two perpendicular lines and each other; use geometry and distance between centres.Explanation:The large circle has a diameter of $8$ cm, so its radius $R = 4$ cm.It touches two perpendicular lines; its centre is at $(4,4)$ from the lines.Let the small circle have radius $r$ cm. It also touches both lines, so its centre is at $(r,r)$.Both circles touch externally, so the distance between centres $= R + r = 4 + r$.Using coordinates: distance $= \sqrt{(4-r)^2 + (4-r)^2} = \sqrt{2}\,(4-r)$.Thus: $\sqrt{2}\,(4-r) = 4 + r$.Expand: $4\sqrt{2} - r\sqrt{2} = 4 + r$.Rearrange: $4\sqrt{2} - 4 = r + r\sqrt{2}$.Factor: $4(\sqrt{2} - 1) = r(1 + \sqrt{2})$.Solve for $r$: $r = \frac{4(\sqrt{2} - 1)}{\sqrt{2} + 1}$.Rationalise: $r = \frac{4(\sqrt{2} - 1)^2}{2 - 1} = 4(3 - 2\sqrt{2})$ cm.Diameter of the small circle $= 2r = 8(3 - 2\sqrt{2})$ cm.Answer:$8(3 - 2\sqrt{2})$ cm, which matches option D.