Given circle :
$$x^2+y^2=25$$
So, center $$O(0,0)$$ and radius $$r=5$$
Given external point :
$$P(5,12)$$
First find
$$OP$$
$$OP=\sqrt{5^2+12^2}$$
$$=\sqrt{25+144}$$
$$=\sqrt{169}$$
$$=13$$
Length of tangent from external point :
$$PA=PB=\sqrt{OP^2-r^2}$$
$$=\sqrt{13^2-5^2}$$
$$=\sqrt{169-25}$$
$$=\sqrt{144}$$
$$=12$$
Now chord of contact $$AB$$ is perpendicular to $$OP$$.
Distance of chord of contact from center is $$\frac{r^2}{OP}$$
Therefore,
distance from center to chord
$$AB$$ is $$\frac{25}{13}$$
Hence perpendicular distance from $$P$$ to chord $$AB$$ is $$13-\frac{25}{13}$$
$$=\frac{169-25}{13}$$
$$=\frac{144}{13}$$
Now length of chord
$$AB$$
is
$$2\sqrt{r^2-d^2}$$
where
$$d=\frac{25}{13}$$
Therefore,
$$AB=2\sqrt{25-\left(\frac{25}{13}\right)^2}$$
$$=2\sqrt{\frac{4225-625}{169}}$$
$$=2\sqrt{\frac{3600}{169}}$$
$$=2\cdot\frac{60}{13}$$
$$=\frac{120}{13}$$
Area of triangle
$$PAB$$
is
$$\frac12\times AB\times \text{height}$$
$$=\frac12\times\frac{120}{13}\times\frac{144}{13}$$
$$=\frac{60\times144}{169}$$
$$=\frac{8640}{169}$$
Final Answer :
$$\frac{8640}{169}$$
91- 6303239042
