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Question

The equation of the circle which passes through the points $$(1, 0), (0, -6)$$ and $$(3, 4)$$ is


A
4x2+4y2+142x+47y+140=0
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B
4x2+4y2142x47y+138=0
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C
4x2+4y2142x+47y+138=0
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D
4x2+4y2+150x49y+138=0
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Solution

The correct option is C $$4x^{2} + 4y^{2} - 142x + 47y + 138 = 0$$

The  general  fromula  of  circle  is

$$ { x }^{ 2 }+{ y }^{ 2 }+Dx+Ey+F=0$$

Putting $$ (1,0)  ;  1+0+D+0+F=0$$

$$ (0,-6);  0+36+0+E+F=0$$

$$ (3,4);  9+16+3D+4E+F=0$$

Solving  the  above  equatio  we  get

$$ D=- \dfrac { 142 }{ 4 }   E= \dfrac { 47 }{ 4 }   F= \dfrac { 138 }{ 4 } $$

Putting  these  value  we  get  equation  of  circle

$$ { 4x }^{ 2 }+4{ y }^{ 2 }-142x+47y+138=0$$

So correct answer will be option C



Mathematics

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