Hyperbola
Intersection of Curves
IIT-JEE 2010 Paper 1
Grade 11
Question:
The circle $x^2 + y^2 - 8x = 0$ and hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ intersect at $A$ and $B$. The equation of the circle with $AB$ as diameter is:
(1) $x^2 + y^2 - 12x + 24 = 0$
(2) $x^2 + y^2 + 12x + 24 = 0$
(3) $x^2 + y^2 - 12x - 24 = 0$
(4) $x^2 + y^2 + 12x - 24 = 0$
Step-by-Step Solution
Key Concept: Find $A$ and $B$: circle $(x - 4)^2 + y^2 = 16$. On hyperbola $y^2 = 4(x^2/9 - 1)$. Substitute: $(x - 4)^2 + 4x^2/9 - 4 = 16 \Rightarrow$ solve for $x$. Midpoint of $AB$ gives centre; use $AB$ as diameter.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)