Let the equation of a circle be x^{2}+y^{2}-6 | vedclass

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(A) The given equation of the circle is $x^{2}+y^{2}-6x-4y+9=0$. Comparing this with the general form $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-3, f=-2, c=

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tangent of the circle

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(A) The given equation of the circle is $x^{2}+y^{2}-6x-4y+9=0$. Comparing this with the general form $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-3, f=-2, c=9$. The center of the circle is $(-g, -f) = (3, 2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{(-3)^{2}+(-2)^{2}-9} = \sqrt{9+4-9} = \sqrt{4} = 2$. To check if the line $4x+3y-8=0$ is a tangent,we calculate the perpendicular distance $d$ from the center $(3, 2)$ to the line: $d = \left|\frac{4(3)+3(2)-8}{\sqrt{4^{2}+3^{2}}}\right| = \left|\frac{12+6-8}{\sqrt{16+9}}\right| = \left|\frac{10}{5}\right| = 2$. Since the perpendicular distance $d$ is equal to the radius $r$ $(d=r=2)$,the line is a tangent to the circle.

Let the equation of a circle be $x^{2}+y^{2}-6x-4y+9=0$. Then the line $4x+3y-8=0$ is a

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