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

4 x^{2} + 4 y^{2} + 142 x + 47 y + 140 = 0

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

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

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4 x^{2} + 4 y^{2} + 150 x - 49 y + 138 = 0

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Solution

The correct option is C $4 x^{2} + 4 y^{2} - 142 x + 47 y + 138 = 0$
The general fromula of circle is
$x^{2} + y^{2} + D x + E y + F = 0$
Putting $(1, 0); 1 + 0 + D + 0 + F = 0$
$(0, - 6); 0 + 36 + 0 + E + F = 0$
$(3, 4); 9 + 16 + 3 D + 4 E + F = 0$
Solving the above equatio we get
$D = - \frac{142}{4} E = \frac{47}{4} F = \frac{138}{4}$
Putting these value we get equation of circle
${4 x}^{2} + 4 y^{2} - 142 x + 47 y + 138 = 0$

So correct answer will be option C

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