The tangents are drawn from the point (4, 5) | vedclass

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(C) The equation of the circle is $x^2 + y^2 - 4x - 2y - 11 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -1$,and $c = -11$

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$8$

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(C) The equation of the circle is $x^2 + y^2 - 4x - 2y - 11 = 0$.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -2$,$f = -1$,and $c = -11$.The center of the circle is $(-g, -f) = (2, 1)$ and the radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{4 + 1 + 11} = \sqrt{16} = 4$.The length of the tangent $L$ from point $(x_1, y_1) = (4, 5)$ is given by $L = \sqrt{x_1^2 + y_1^2 - 4x_1 - 2y_1 - 11}$.$L = \sqrt{4^2 + 5^2 - 4(4) - 2(5) - 11} = \sqrt{16 + 25 - 16 - 10 - 11} = \sqrt{4} = 2$.The quadrilateral formed by the two tangents,the two radii,and the center is composed of two congruent right-angled triangles.The area of the quadrilateral is $2 	imes (\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}) = 2 	imes (\frac{1}{2} 	imes L 	imes r) = L 	imes r$.Area $= 2 	imes 4 = 8 	ext{ sq. units}$.

The tangents are drawn from the point $(4, 5)$ to the circle $x^2 + y^2 - 4x - 2y - 11 = 0$. The area of the quadrilateral formed by these tangents and the radii is .............. $sq. \text{ units}$.

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