If the two lines x - y -6 and x -2 y -4 are the diameters of the circle which passes through 2-6- then its equation is A- x2-y2-16 x-4 y-32-0B- x2-y2-16 x-4 y-23-0C- x 2- y 2-16 x -4 y -32-0D- x2-y2-16 x-4 y-23-0

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Definition of Circle

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Question

If the two lines $x + y = 6$ and $x + 2 y = 4$ are the diameters of the circle which passes through $(2, 6)$ , then its equation is

x^{2} + y^{2} - 16 x + 4 y + 32 = 0

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x^{2} + y^{2} - 16 x + 4 y + 23 = 0

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x^{2} + y^{2} - 16 x + 4 y - 32 = 0

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x^{2} + y^{2} - 16 x + 4 y - 23 = 0

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Solution

The correct option is C $x^{2} + y^{2} - 16 x + 4 y - 32 = 0$
Given lines:
$x + y = 6, x + 2 y = 4$
Point of intersection of these lines is the centre of the circle
$C \equiv (8, - 2)$

Radius is the distance between $(8, - 2)$ and $(2, 6)$
$r = \sqrt{36 + 64} = 10$

So, the equation of required circle is
$(x - 8)^{2} + (y + 2)^{2} = 10^{2}$
$∴ x^{2} + y^{2} - 16 x + 4 y - 32 = 0$

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If the two lines x+y=6 and x+2y=4 are the diameters of the circle which passes through (2,6), then its equation is

The correct option is C x2+y2−16x+4y−32=0Given lines: x+y=6, x+2y=4 Point of intersection of these lines is the centre of the circle C≡(8,−2) Radius is the distance between (8,−2) and (2,6) r=√36+64=10 So, the equation of required circle is (x−8)2+(y+2)2=102 ∴x2+y2−16x+4y−32=0

If the two lines $x + y = 6$ and $x + 2 y = 4$ are the diameters of the circle which passes through $(2, 6)$ , then its equation is