Question:medium

The diameters of a circle are along $2x+y-7=0$ and $x+3y-11=0$. Then, the equation of this circle, which also passes through (5, 7), is

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The center of a circle is the intersection of any two of its diameters.

Updated On: Apr 10, 2026

$x^{2}+y^{2}-4x-6y-16=0$
$x^{2}+y^{2}-4x-6y-20=0$
$x^{2}+y^{2}-4x-6y-12=0$
$x^{2}+y^{2}+4x+6y-12=0$

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The Correct Option is C

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The diameters of a circle are along $2x+y-7=0$ and $x+3y-11=0$. Then, the equation of this circle, which also passes through (5, 7), is

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The Correct Option is C

Solution and Explanation

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