Circles $C_1$, $C_2$, and $C_3$, with centres $O_1$, $O_2$, and $O_3$, and radii $r_1$, $r_2$, and $r_3$ respectively, touch each other as shown in the figure. Given $r_1 = 2\text{ cm}$, $r_2 = 1\text{ cm}$ and the angle $\angle O_1O_3O_2 = 90^\circ$, $r_3 = \text{_____ cm}.$
Options:
(A) $\frac{1}{2}(-3 + \sqrt{17})$
(B) $\frac{1}{2}(3 + \sqrt{17})$
(C) $\frac{1}{2}(-2 + \sqrt{17})$
(D) $\frac{1}{2}(-3 + 2\sqrt{17})$
Official Answer Key:
A
Step-by-Step Solution & Explanation
For circles touching externally, the distance between centres equals the sum of the radii:
O₁O₃ = r₁ + r₃ = 2 + r₃
O₂O₃ = r₂ + r₃ = 1 + r₃
O₁O₂ = r₁ + r₂ = 3
Since ∠O₁O₃O₂ = 90°, triangle O₁O₃O₂ is right-angled at O₃, so by Pythagoras:
(2 + r₃)² + (1 + r₃)² = 3²
(4 + 4r₃ + r₃²) + (1 + 2r₃ + r₃²) = 9
2r₃² + 6r₃ − 4 = 0
r₃² + 3r₃ − 2 = 0
r₃ = [−3 ± √(9 + 8)]/2 = ½(−3 ± √17)
Radius must be positive ⇒ r₃ = ½(−3 + √17) ≈ 0.562 cm.
SECTION 1: ENGINEERING MATHEMATICS
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