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(C) The equation of the circle is $x^2+y^2-4x-2y-11=0$.Rewriting this in standard form: $(x-2)^2+(y-1)^2 = 11+4+1 = 16$.Thus,the center $C$ is $(2,1)$

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(C) The equation of the circle is $x^2+y^2-4x-2y-11=0$.Rewriting this in standard form: $(x-2)^2+(y-1)^2 = 11+4+1 = 16$.Thus,the center $C$ is $(2,1)$ and the radius $r = \sqrt{16} = 4$.The point $P$ is $(4,5)$.The length of the tangent $AP$ from $P(4,5)$ to the circle is given by $\sqrt{S_1} = \sqrt{4^2+5^2-4(4)-2(5)-11} = \sqrt{16+25-16-10-11} = \sqrt{4} = 2$.The quadrilateral formed is $PACB$,where $A$ and $B$ are points of contact.This quadrilateral consists of two congruent right-angled triangles $	riangle PAC$ and $	riangle PBC$,both right-angled at $A$ and $B$ respectively.The area of $	riangle PAC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AC 	imes AP = \frac{1}{2} 	imes 4 	imes 2 = 4$.The total area of the quadrilateral $PACB = 2 	imes 	ext{Area}(	riangle PAC) = 2 	imes 4 = 8 	ext{ sq units}$.

The area of the quadrilateral formed by the tangents from the point $(4,5)$ to the circle $x^2+y^2-4x-2y-11=0$,and the pair of radii joining the points of contact of these tangents is

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