Chord of Contact for a Circle at Origin

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 :

$$\frac{8640}{169}$$