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Question: Area of quadrilateral formed by a pair of tangents from the point \(\left( 4,5 \right)\) to the circ...

Area of quadrilateral formed by a pair of tangents from the point (4,5)\left( 4,5 \right) to the circle x2+y2−4x−2y−11=0{{x}^{2}}+{{y}^{2}}-4x-2y-11=0 and a pair of radii is
A. 12 sq-unit
B. 4 sq-unit
C. 6 sq-unit
D. 8 sq-unit

Explanation

Solution

First let us draw a diagram of circle and a pair of tangents from the given point. Then, we compare the equation of the circle given with the general equation of the circle and find the center and radius of the circle. Then, calculate the area of quadrilateral formed with the help of formula Area=radius×length of tangent\text{Area=radius}\times \text{length of tangent}

Complete step-by-step answer :
We have been given that a quadrilateral is formed by a pair of tangents from the point (4,5)\left( 4,5 \right) to the circle x2+y2−4x−2y−11=0{{x}^{2}}+{{y}^{2}}-4x-2y-11=0 and a pair of radii.
We have to find the area of quadrilateral formed.

Let us assume that AA is the center of the circle and AB&ACAB\And AC are radius of the circle. Tangents are drawn from the point DD.
If x2+y2+2gx+2fy+c=0{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 is the general equation of the circle then, the center of circle is given by (−g,−f)\left( -g,-f \right) and radius of circle is given by r=g2+f2−cr=\sqrt{{{g}^{2}}+{{f}^{2}}-c}.
We have an equation of circle x2+y2−4x−2y−11=0{{x}^{2}}+{{y}^{2}}-4x-2y-11=0, when we compare it with the general equation, we get center of circle will be A(2,1)A\left( 2,1 \right) and radius of circle will be
AC=AB=g2+f2−c AC=AB=(−2)2+(−1)2−(−11) AC=AB=4+1+11 AC=AB=16 AC=AB=4 \begin{aligned} & AC=AB=\sqrt{{{g}^{2}}+{{f}^{2}}-c} \\\ & AC=AB=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -1 \right)}^{2}}-\left( -11 \right)} \\\ & AC=AB=\sqrt{4+1+11} \\\ & AC=AB=\sqrt{16} \\\ & AC=AB=4 \\\ \end{aligned}
Now, we know that area of quadrilateral formed will be radius×length of tangent\text{radius}\times \text{length of tangent}
Length of tangent will be x2+y2−4x−2y−11\sqrt{{{x}^{2}}+{{y}^{2}}-4x-2y-11} , when we put the value of point(4,5)\left( 4,5 \right) given in the question, we get
Length of tangents
=42+52−4×4−2×5−11 =16+25−16−10−11 =4 =2 \begin{aligned} & =\sqrt{{{4}^{2}}+{{5}^{2}}-4\times 4-2\times 5-11} \\\ & =\sqrt{16+25-16-10-11} \\\ & =\sqrt{4} \\\ & =2 \\\ \end{aligned}
Now, the area of quadrilateral formed will be 4×2=84\times 2=8
So, the required area is 8 sq-unit.
Option D is the correct answer.

Note : The lengths of tangents drawn to a circle from the same external point are equal. The length of tangents is also found by using the distance formula when we have coordinates of two points. Also we could calculate the area of two triangles formed by joining the centre to the outside point and add them to get the quadrilateral area.