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

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

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

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

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

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Solution

The correct option is C
$x^{2} + y^{2} - 4 x - 6 y - 12 = 0$

C = (2, 3), r = 5

$∴ The equation of circle is x^{2} + y^{2} - 4 x - 6 y - 12 = 0$

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x^{2} + y^{2} - 4 x - 6 y = 0

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Tangents are drawn from the point P(1, 8) to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the point A and B, then equation of the circumcircle of the triangle PAB is

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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

The correct option is C x2+y2−4x−6y−12=0 C = (2, 3), r = 5 ∴The equation of circle is x2+y2−4x−6y−12=0

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

The correct option is C
$x^{2} + y^{2} - 4 x - 6 y - 12 = 0$

C = (2, 3), r = 5

$∴ The equation of circle is x^{2} + y^{2} - 4 x - 6 y - 12 = 0$