The equation of the circle passing through the point (0, 0) and the points where the straight line 3x - 4y = 12 meets the axes of coordinates is.

x^{2} + y^{2} + 4 x - 3 y = 0

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x^{2} + y^{2} - 4 x - 3 y = 0

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x^{2} + y^{2} - 4 x + 3 y = 0

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Solution

The correct option is C $x^{2} + y^{2} - 4 x + 3 y = 0$
Putting $x = 0$ in $3 x - 4 y = 12 :$
$\Rightarrow$ $3 (0) - 4 y = 12$
$∴$ $y = - 3$
Putting $y = 0$ in $3 x - 4 y = 12 :$
$\Rightarrow$ $3 x - 4 (0) = 12$
$∴$ $x = 4$
Thus, the line $3 x + 4 y = 12$ meets the axes of coordinates at points $A (0, - 3)$ and $B (4, 0) .$
The equation of the circle with AB as the diameter is
$(x - 0) (x - 4) + (y + 3) (y - 0) = 0$
$\Rightarrow$ $x^{2} - 4 x + y^{2} + 3 y = 0$
$\Rightarrow$ $x^{2} + y^{2} - 4 x + 3 y = 0$

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