Circle
Equation of Circle
IIT-JEE 2011 Paper 2
Grade 11
Question:
The circle passing through $(-1, 0)$ and touching the $y$-axis at $(0, 2)$ also passes through the point:
(1) $(-\frac{3}{2}, 0)$
(2) $(-\frac{5}{2}, 2)$
(3) $(-\frac{3}{2}, \frac{5}{2})$
(4) $(-4, 0)$
Step-by-Step Solution
Key Concept: A circle touching the $y$-axis at $(0, 2)$ has centre $(-a, 2)$ and radius $a$. Substituting $(-1, 0)$: $(-1 + a)^2 + 4 = a^2$ gives $a = 5/2$. Then verify $(-4, 0)$ lies on $(x + \frac{5}{2})^2 + (y - 2)^2 = \frac{25}{4}$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (4)