If the circle x^2 + y^2 - 6x - 8y + (25 - a^2 | vedclass

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(B) The given equation of the circle is $x^2 + y^2 - 6x - 8y + (25 - a^2) = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we

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$\pm 4$

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(B) The given equation of the circle is $x^2 + y^2 - 6x - 8y + (25 - a^2) = 0$.Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -3$,$f = -4$,and $c = 25 - a^2$.The center of the circle is $(-g, -f) = (3, 4)$ and the radius $r$ is given by $\sqrt{g^2 + f^2 - c}$.$r = \sqrt{(-3)^2 + (-4)^2 - (25 - a^2)} = \sqrt{9 + 16 - 25 + a^2} = \sqrt{a^2} = |a|$.Since the circle touches the $x$-axis,the radius must be equal to the absolute value of the $y$-coordinate of the center.Thus,$|a| = |4|$,which implies $a = \pm 4$.

If the circle $x^2 + y^2 - 6x - 8y + (25 - a^2) = 0$ touches the $x$-axis,then $a$ equals

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