As shown in the figure, circle C1 with center O1 and radius r1 touches the square
VWXY at points P and Q while circle C2 with center O2 and radius r2 touches the
square VWXY at points R and S. The two circles touch each other at T.
Given r1 = 1 cm and VY̅̅̅̅ = VW̅̅̅̅̅ = 4 cm, r2 = _____ cm.
Options:
(A) 4 - 3√2
(B) 1 + 2√2
(C) 7 - 4√2
(D) 5 + 3√2
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.10)**
**1. Intuitive Physical Setup & Given Parameters**:
- Square $VWXY$ has side length $s = 4 ext{ cm}$.
- Circle $C_1$ (center $O_1$, radius $r_1 = 1 ext{ cm}$) touches two adjacent sides of the square at $P$ and $Q$ near corner $V$.
- Circle $C_2$ (center $O_2$, radius $r_2$) touches the other two adjacent sides at $R$ and $S$ near corner $X$.
- The two circles touch each other externally at point $T$.
- We need to find $r_2$.
**2. Step-by-Step Mathematical Derivation**:
1. **Geometry Along the Main Diagonal**:
- By symmetry, both circle centers $O_1$ and $O_2$ lie on the main diagonal $VX$ of the square.
- The total length of the diagonal of the square of side $s = 4 ext{ cm}$ is:
$VX = s\sqrt{2} = 4\sqrt{2}\text{ cm}$
2. **Decomposing the Diagonal Segment**:
- The distance from corner $V$ to center $O_1$ is the hypotenuse of an isosceles right triangle with legs $r_1$:
$VO_1 = r_1\sqrt{2}$
- The distance from center $O_2$ to opposite corner $X$ is the hypotenuse with legs $r_2$:
$O_2X = r_2\sqrt{2}$
- Since the circles touch externally at $T$, the distance between their centers is:
$O_1O_2 = r_1 + r_2$
- The diagonal $VX$ is the sum of these collinear segments:
$VX = VO_1 + O_1O_2 + O_2X = r_1\sqrt{2} + (r_1 + r_2) + r_2\sqrt{2}$
3. **Formulate and Solve for $r_2$**:
$r_1(\sqrt{2} + 1) + r_2(\sqrt{2} + 1) = 4\sqrt{2}$
$(r_1 + r_2)(\sqrt{2} + 1) = 4\sqrt{2}$
$r_1 + r_2 = \frac{4\sqrt{2}}{\sqrt{2} + 1} = \frac{4\sqrt{2}(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = 4(2 - \sqrt{2}) = 8 - 4\sqrt{2}$
Given $r_1 = 1\text{ cm}$:
$r_2 = (8 - 4\sqrt{2}) - 1 = 7 - 4\sqrt{2}\text{ cm}$
**3. Comprehensive Options & Distractor Analysis**:
- **(A) 4 - 3√2**: ❌ **INCORRECT**. Numerical value $\approx -0.24$, which is negative and unphysical.
- **(B) 1 + 2√2**: ❌ **INCORRECT**. Numerical value $\approx 3.83\text{ cm}$, too large since $r_1 + r_2$ would exceed the allowable diagonal space.
- **(C) 7 - 4√2**: ✅ **CORRECT**. $7 - 4(1.4142) = 7 - 5.6568 = 1.3432\text{ cm}$, perfectly satisfying the geometric constraint.
- **(D) 5 + 3√2**: ❌ **INCORRECT**. Far exceeds the dimensions of the square.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **C**
- **Exam Takeaway**: Circles inscribed against the corners of a polygon have centers lying on angle bisectors. The diagonal projection equation $(r_1 + r_2)(\sqrt{2} + 1) = s\sqrt{2}$ yields instant solutions via algebraic rationalization.
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