Question

Area of quadrilateral PACB, is

• A. $4$
• B. $3$
• C. $2$
• D. $8$

Answer 1

Answer: (A)

$4$
Given equation of circle $x^2+y^2-4x-6y-3=0$, which can be simplified to
$(x-2{)}^2+(y-3{)}^2=4^2$
It's a circle of radius 4 with center at (2,3).

Two tangents are drawn from point P (1, -1), which meets the above circle at and .

Length of CP = $\sqrt{{(2-1)}^2+{(3+1)}^2}=\sqrt{17}$
Length of AC = radius of circle = 4
$△ACP$ is a right angled triangle, Using Pythagorus theorem, Length of AP = $\sqrt{C{P}^2-A{C}^2}=\sqrt{{\sqrt{17}}^2-4^2}=1$
Area of $△CBP$ is equal to Area of $△ACP$
$\therefore$ Area of quadrilateral PACB = 2*Area of $△ACP=2\ast \frac{1}{2}\ast AP\ast AC=2\ast \frac{1}{2}\ast 4\ast 1=4$
Hence, Option A.