Question

Let $x^2+y^2-4x-2y-11=0$ be a circle. A pair of tangents from the point $(4,5)$ with a pair of radii from a quadrilateral of area....

Answer 1

We have given from the question

$x^2+y^2-4x-2y-11=0$

${\left(x-2\right)}^2+{\left(y-1\right)}^2={\left(4\right)}^2$

Now,

$I=\sqrt{S_1}$

$=\sqrt{{\left(4\right)}^2+{\left(5\right)}^2-4\left(4\right)-2\left(5\right)-11}$

$=\sqrt{41-16-10-11}$

$=\sqrt{4}$

$=2$

Now,

Area of quadrilateral $ABCD=2\times \frac{1}{2}\times 2\times 4$

$=8$

Hence, the required areal of triangle is $8$.