The graph of $x^{2} - 4 x + y^{2} + 6 y = 0$ in the xy-plane is a circle. What is the radius of the circle?

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Solution

Given,
$x^{2} - 4 x + y^{2} + 6 y = 0$
$x^{2} - 4 x + 4 - 4 + y^{2} + 6 y + 9 - 9 = 0$
$(x - 2)^{2} + (y + 3)^{2} - 13 = 0$
$(x - 2)^{2} + (y + 3)^{2} = 13$
Equation of circle ,
$(x - h)^{2} + (y - k)^{2} = r^{2}$
by comparing
$r^{2} = 13$
$r =$ $\sqrt{13}$
so , radius of circle is $\sqrt{13}$

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