lf the area of the quadrilateral formed by the tangent from the origin to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}+6\mathrm{x}- 1 \mathrm{0}\mathrm{y}+\mathrm{c}=0$ and the pair of radii at the points of contact of these tangents to the circle is $8$ square units, then $c$ is a root of the equation

For a circle x^2 + y^2 + 2gx + 2fy + c = 0, the radius r = sqrt(g^2 + f^2 - c). Here g=3, f=-5, so r = sqrt(9 + 25 - c) = sqrt(34 - c). The distance from origin to center is d = sqrt(3^2 + (-5)^2) = sqrt(34). The area of the quadrilateral formed by tangents and radii is r * sqrt(d^2 - r^2). Given area = 8, so r * sqrt(34 - r^2) = 8. Squaring gives r^2 * (34 - r^2) = 64. Substituting r^2 = 34 - c, we get (34 - c) * (34 - (34 - c)) = 64, which simplifies to (34 - c) * c = 64, or c^2 - 34c + 64 = 0.