Let \[P({{x}_{1}},{{y}_{1}})\] be a point lying inside the circle \[S:{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]. If the tangents from P to the circle \[S=0\] at A and B, then AB is called the chord of contact.
The equation of the chord of contact is \[x{{x}_{1}}+y{{y}_{1}}++g(x+{{x}_{1}})+f(y+{{y}_{1}})+c=0\]. Let the tangents PA and PB are drawn from \[P(0,-2)\] to the circle \[{{x}^{2}}+{{y}^{2}}+2x-4y=0\].
The area of a quadrilateral PACB, where C is the centre of the circle, is
(a) \[2\sqrt{15}\] (b) \[3\sqrt{15}\] (c) \[4\sqrt{15}\] (d) \[5\sqrt{15}\]

Hint:We know that the equation of circle \[S=0\] or \[S:{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] can be used to, find the centre of circle and radius directly. The co-ordinates of the centre is \[(-g,-f)\] and the radius given as \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]. And the length of a tangent drawn on circle \[S=0\] from a point \[P(0,-2)\] outside the circle is given by \[\sqrt{{{S}_{1}}}\] or \[\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}+2g{{x}_{1}}+2f{{y}_{1}}+c}\]. And take area of triangle \[PAC\] using – \[\text{Area of }\vartriangle \text{=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ Base }\!\!\times\!\!\text{ Height}\].

Complete step by step answer:
First we need to derive equation of a circle as –
A circle is the locus of all points in the plane which has a fixed constant distance from a fixed point called centre. Where we assume co-ordinates of centre of the circle (O) as \[O(h,k)\] and radius r and a variable point \[P(x,y)\],
 Its equation is \[\sqrt{{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}}=r\]
When we square both sides and expand brackets,
\[{{x}^{2}}+{{y}^{2}}-2hx-2ky+{{h}^{2}}+{{k}^{2}}={{r}^{2}}\]
Let us put \[g=-h,f=-k,c={{h}^{2}}+{{k}^{2}}-{{r}^{2}}\],
\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]
Hence we obtained above result and in addition to it we obtained the following results,
\[r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]
\[O(-g,-f)\]
As we know that the general equation of a circle can be given \[S:{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] or simply, \[S=0\].
The centre of the circle \[S=0\] is given as \[(-g,-f)\] and the radius of the circle given by formula \[r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\].
If S: \[S:{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] represents a general equation of circle, then the power of a point \[P({{x}_{1}},{{y}_{1}})\] is given as
\[{{S}_{1}}=S({{x}_{1}},{{y}_{1}})\] whose sign indicates the position of point \[P(x,y)\] with respect to the circle \[S=0\]
\[\text{Now Applying Area of Parallelogram = 2}\times \text{Area of }\vartriangle \text{PAC}\]
\[=2\times \dfrac{1}{2}\times r\times \sqrt{{{S}_{1}}}\]
\[=r\sqrt{{{S}_{1}}}\]
\[=\sqrt{5}\times \mathrm{2}\sqrt{3}\]
\[=2\sqrt{15}\]
Hence the correct option is \[A\].

Note:
The student must be aware of the general equation of a circle i.e. \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] and the ways to derive the formula for it, in addition to it the student must know the way to derive the formula for the length of the tangent from an external point to a particular circle also the method to find the radius of a general circle must be known. The common mistakes committed by students are forgetting the formula for finding the radius, center, chord, etc. of a general circle.