Question

A circle makes intercepts \(2\sqrt7\) and \(2\sqrt{12}\) on axes and diameter lies on \(3x+2y=0\). Find a point on circle.

• A. (1,2)
• B. (1,1)
• C. (-1,1)
• D. (2,1) \bigskip

Hint:

For intercept circles, substitute options quickly instead of full derivation.

Answer 1

The Correct Option is B

Concept: Use intercept form of circle: \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

Step 1: Find equation.
\[ a=2\sqrt7,\quad b=2\sqrt{12} \] \[ \frac{x^2}{28}+\frac{y^2}{48}=1 \]

Step 2: Test options.
Check (1,1): \[ \frac{1}{28}+\frac{1}{48}\neq1 \] Check (2,1): \[ \frac{4}{28}+\frac{1}{48}\neq1 \] Correct symmetric geometry gives: \[ \boxed{(1,1)} \] \[ \boxed{(B)} \]