If the radius of a circle x^{2}+y^{2}-4x+6y-k | vedclass

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(D) The general equation of a circle is $x^{2}+y^{2}+2gx+2fy+c=0$,where the radius $r = \sqrt{g^{2}+f^{2}-c}$.Comparing the given equation $x^{2}+y^{2

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$12$

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(D) The general equation of a circle is $x^{2}+y^{2}+2gx+2fy+c=0$,where the radius $r = \sqrt{g^{2}+f^{2}-c}$.Comparing the given equation $x^{2}+y^{2}-4x+6y-k=0$ with the general form:$2g = -4 \Rightarrow g = -2$$2f = 6 \Rightarrow f = 3$$c = -k$Given the radius $r = 5$,we have:$5 = \sqrt{(-2)^{2} + (3)^{2} - (-k)}$$5 = \sqrt{4 + 9 + k}$Squaring both sides:$25 = 13 + k$$k = 25 - 13 = 12$

If the radius of a circle $x^{2}+y^{2}-4x+6y-k=0$ is $5$,then $k=$

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