CAT 2018

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Circles — Perpendicular from centre bisects a chord
• Formula: $$ r^2 = d^2 + \left(\frac{\text{chord length}}{2}\right)^2 $$

where $$r$$ is the radius and $$d$$ is the perpendicular distance from the centre to the chord.

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2. Calculation

Let the two parallel chords be $$AB = 6$$ cm and $$CD = 4$$ cm, both on the same side of the diameter. The distance between them is $$MN = 1$$ cm, where $$M$$ and $$N$$ are the feet of the perpendiculars from the centre $$O$$ to chords $$AB$$ and $$CD$$ respectively.

Since the perpendicular from the centre bisects a chord: $$AM = \frac{6}{2} = 3 \text{ cm}, \quad CN = \frac{4}{2} = 2 \text{ cm}$$

Let $$OM = x$$ cm (perpendicular distance from centre to the longer chord $$AB$$). Since both chords are on the same side and $$CD$$ is farther from the centre (being shorter), the distance from centre to $$CD$$ is $$ON = x + 1$$.

Using the Pythagorean theorem in $$\triangle AMO$$: $$r^2 = AM^2 + OM^2 = 3^2 + x^2 = 9 + x^2 \quad \cdots (1)$$

Using the Pythagorean theorem in $$\triangle CNO$$: $$r^2 = CN^2 + ON^2 = 2^2 + (x+1)^2 = 4 + x^2 + 2x + 1 \quad \cdots (2)$$

Equating (1) and (2): $$9 + x^2 = 4 + x^2 + 2x + 1$$ $$9 = 5 + 2x$$ $$2x = 4 \implies x = 2 \text{ cm}$$

Substituting $$x = 2$$ back into equation (1): $$r^2 = 9 + 4 = 13$$ $$r = \sqrt{13} \text{ cm}$$

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3. Solution

Answer = Option A ✅

The radius of the circle is $$\sqrt{13}$$ cm.