Question:
The equation of circle which passes through the origin and cuts off intercepts 5 and 6 from the positive parts of the axes respectively, is \( \left(x - \frac{5}{2}\right)^2 + (y - 3)^2 = \lambda \), where \( \lambda \) is
The equation of circle which passes through the origin and cuts off intercepts 5 and 6 from the positive parts of the axes respectively, is \( \left(x - \frac{5}{2}\right)^2 + (y - 3)^2 = \lambda \), where \( \lambda \) is
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When circle passes through a point, substitute it into the equation to find radius or constant.
Updated On: Apr 23, 2026
- \( \frac{61}{4} \)
- \( \frac{6}{4} \)
- \( \frac{1}{4} \)
- \( 0 \)
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The Correct Option is A
Solution and Explanation
Concept:
Intercept form of circle gives center at midpoint of intercepts.
Step 1: Intercepts are 5 and 6. So center: \[ \left(\frac{5}{2}, 3\right) \]
Step 2: Use origin \( (0,0) \) to find radius: \[ \lambda = \left(0 - \frac{5}{2}\right)^2 + (0 - 3)^2 \] \[ = \frac{25}{4} + 9 = \frac{25}{4} + \frac{36}{4} = \frac{61}{4} \] Final Answer: \[ \frac{61}{4} \]
Step 1: Intercepts are 5 and 6. So center: \[ \left(\frac{5}{2}, 3\right) \]
Step 2: Use origin \( (0,0) \) to find radius: \[ \lambda = \left(0 - \frac{5}{2}\right)^2 + (0 - 3)^2 \] \[ = \frac{25}{4} + 9 = \frac{25}{4} + \frac{36}{4} = \frac{61}{4} \] Final Answer: \[ \frac{61}{4} \]
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