If the radius of the circle x^2 + y^2 - 18x + | vedclass

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(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Comparing the given equation $x^2 + y^2 - 18x + 12y + k = 0$ with the general

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$-4$

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(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Comparing the given equation $x^2 + y^2 - 18x + 12y + k = 0$ with the general form,we get $2g = -18 \implies g = -9$ and $2f = 12 \implies f = 6$. The constant term is $c = k$. The radius $r$ of the circle is given by the formula $r = \sqrt{g^2 + f^2 - c}$. Given $r = 11$,we have $11 = \sqrt{(-9)^2 + (6)^2 - k}$. Squaring both sides,$121 = 81 + 36 - k$. $121 = 117 - k$. $k = 117 - 121 = -4$.

If the radius of the circle $x^2 + y^2 - 18x + 12y + k = 0$ is $11$,then $k = $

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