Find the equation of the circle with center on x + y = 4 and 5x + 2y + 1 = 0 and having a radius of 3

x^{2} + y^{2} + 6 x - 16 y + 64 = 0

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x^{2} + y^{2} + 8 x - 14 y + 25 = 0

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x^{2} + y^{2} + 6 x - 14 y + 49 = 0

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x^{2} + y^{2} + 6 x - 14 y + 36 = 0

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none of these

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Solution

The correct option is B $x^{2} + y^{2} + 6 x - 14 y + 49 = 0$
Given,
Centre of circle is on $x + y = 4$ and $5 x + 2 y + 1 = 0$
Thus,
$x + y = 4$
$=> y = 4 - x$ (i)
$5 x + 2 y + 1 = 0$ (ii)
$=> 5 x + 2 (4 - x) + 1 = 0$
$=> 5 x + 8 - 2 x + 1 = 0$
$=> 3 x + 9 = 0$
$=> x = - 3$
$∴ y = 4 - (- 3)$
$= 7$
$∴$ Centre of circle is $(- 3, 7)$
Radius of circle= $3$ (given)
$∴$ Equation of circle is
$(x - (- 3))^{2} + (y - 7)^{2} = 3^{2}$
$=> (x + 3)^{2} + (y - 7)^{2} = 9$
$=> (x^{2} + 6 x + 9) + (y^{2} - 14 y + 49) - 9 = 0$
$=> x^{2} + 6 x + y^{2} - 14 y + 49 = 0$

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