The equation of the circle passing through the points 4-1-6-5 whose centre lies on the line 4 x-y-16-0 isA- x2-y2-6 x-8 y-15-0B- x2-y2-10 x-6 y-29-0C- x2-y2-x-13-0D- x2-y2-4 y-12-0

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Question

The equation of the circle passing through the points (4,1),(6,5) whose centre lies on the line 4x+y-16=0 is

x²+y²-10x-6y+29=0

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x²+y²-x-13=0

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x²+y²-6x-8y+15=0

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x²+y²+4y-12=0

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Solution

The correct option is C
x²+y²-6x-8y+15=0

$A (4, 1)$
$B (6, 5)$
Centre lies on the line $4 x + y - 16 = 0$
S centre $C = (x, 16 - 4 x)$
$A C = B C$
$\sqrt{(x - 4)^{2} + (y - 1)^{2}} = \sqrt{(x - 6)^{2} + (y - 5)^{2}}$
$(x - 4)^{2} + (y - 1)^{2} = (x - 6)^{2} + (y - 5)^{2}$
Solving we get $(x, y) = (3, 4),$
$A C = \sqrt{10}$
$(x - 3)^{2} + (y - 4)^{2} = 10$

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The equation of the circle passing through the points (4,1),(6,5) whose centre lies on the line 4x+y-16=0 is

The correct option is C x 2+y2-6x-8y+15=0 A(4,1) B(6,5) Centre lies on the line 4x+y−16=0 S centre C=(x,16−4x) AC=BC √(x−4)2+(y−1)2=√(x−6)2+(y−5)2 (x−4)2+(y−1)2=(x−6)2+(y−5)2 Solving we get (x,y)=(3,4), AC=√10 (x−3)2+(y−4)2=10