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Question 9
Question Statement:
A circle with center at (x, y) = (0.5, 0) and radius = 0.5 intersects with another
circle with center at (x, y) = (1, 1) and radius = 1 at two points. One of the points
of intersection (x, y) is:
Options:
(A) (0, 0)
(B) (0.2, 0.4)
(C) (0.5, 0.5)
(D) (1, 2)
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.9)**
**1. Coordinate Geometry Formulation**:
- **Circle 1**: Center $(x_1, y_1) = (0.5, 0)$, radius $R_1 = 0.5$:
$(x - 0.5)^2 + y^2 = (0.5)^2 \implies x^2 - x + 0.25 + y^2 = 0.25 \implies x^2 + y^2 = x$
- **Circle 2**: Center $(x_2, y_2) = (1, 1)$, radius $R_2 = 1$:
$(x - 1)^2 + (y - 1)^2 = 1^2 \implies x^2 - 2x + 1 + y^2 - 2y + 1 = 1 \implies x^2 + y^2 - 2x - 2y + 1 = 0$
**2. Step-by-Step Intersection Derivation**:
- Step 1: Substitute $x^2 + y^2 = x$ into Circle 2 equation:
$x - 2x - 2y + 1 = 0 \implies -x - 2y + 1 = 0 \implies x = 1 - 2y$
- Step 2: Substitute $x = 1 - 2y$ back into $x^2 + y^2 = x$:
$(1 - 2y)^2 + y^2 = 1 - 2y$
$1 - 4y + 4y^2 + y^2 = 1 - 2y$
$5y^2 - 2y = 0 \implies y(5y - 2) = 0$
- Step 3: Solve for intersection roots:
- Root 1: $y = 0 \implies x = 1 - 2(0) = 1 \implies (1, 0)$
- Root 2: $y = \frac{2}{5} = 0.4 \implies x = 1 - 2(0.4) = 1 - 0.8 = 0.2 \implies (0.2, 0.4)$
- The point $(0.2, 0.4)$ is explicitly listed in Option (B).
**3. Official Verification**:
The verified official IIT answer is **Option (B)** (`(0.2, 0.4)`).
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