Question:
The normal at the point (3, 4) on a circle cuts the circle at the point (-1, -2). Then the equation of the circle is
The normal at the point (3, 4) on a circle cuts the circle at the point (-1, -2). Then the equation of the circle is
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Diameter ends give circle equation: \((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\).
Updated On: Apr 7, 2026
- \(x^2 + y^2 + 2x - 2y - 13 = 0\)
- \(x^2 + y^2 - 2x - 2y - 11 = 0\)
- \(x^2 + y^2 - 2x + 2y + 12 = 0\)
- \(x^2 + y^2 - 2x - 2y + 14 = 0\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Normal passes through centre, so (3,4) and (-1,-2) are ends of diameter.
Step 2: Detailed Explanation:
Equation of circle with ends of diameter:
\((x - 3)(x + 1) + (y - 4)(y + 2) = 0\)
\(x^2 - 2x - 3 + y^2 - 2y - 8 = 0\)
\(x^2 + y^2 - 2x - 2y - 11 = 0\)
Step 3: Final Answer:
\(x^2 + y^2 - 2x - 2y - 11 = 0\).
Normal passes through centre, so (3,4) and (-1,-2) are ends of diameter.
Step 2: Detailed Explanation:
Equation of circle with ends of diameter:
\((x - 3)(x + 1) + (y - 4)(y + 2) = 0\)
\(x^2 - 2x - 3 + y^2 - 2y - 8 = 0\)
\(x^2 + y^2 - 2x - 2y - 11 = 0\)
Step 3: Final Answer:
\(x^2 + y^2 - 2x - 2y - 11 = 0\).
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