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If the length of tangent drawn from an external point P to the circle of radius $r$ is $l$ , then prove that area of triangle form by pair of tangent and its chord of contact is $\displaystyle\frac{rl^3}{r^2+l^2}$

I have the solution of this which states: External point P, centre C and one point of contact is A. Let $\theta$ is the angle formed at P and $\angle APC$.

It is given that area =$\frac{1}{2} 2r\cos\theta \cdot l \cos\theta$

My question how come it is $r\cos\theta$ as the point of intersection of chord and line PC is not equal to $r$ (radius).

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We have Area =$2\frac{1}{2} l \sin(\theta)l \cos(\theta)$. However by considering triangle $APC$, we know $\tan(\theta)=\frac{r}{l}$. Hence Area =$2\frac{1}{2} \frac{r}{\tan(\theta)} \sin(\theta)l \cos(\theta)=rl\cos^2(\theta)$ Hence you get the answer in solution.

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Let $A'$ be the other point of contact.

The triangle $APC$ is right. $A'A$ and $PC$ are perpendicular. Let $H$ be the intersection of $A'A$ and $PC$. Then the triangles $ACH$, $AHP$ and $APC$ are similar. Thus, the angle $\angle APC$ is equal to $\angle CAH$. Then, the chord $A'A$ is $2r\cos\theta$. Now, use the similarity:

$$\frac{PH}{AH}=\frac{AP}{AC}$$ that is, $$\frac{PH}{r\cos\theta}=\frac lr$$ and then $$PH=l\cos\theta$$

Now we can compute the area: $$[AA'P]=\frac{AA'\cdot HP}2=rl\cos^2\theta$$ but $$\cos\theta=\frac {AP}{CP}=\frac l{\sqrt{l^2+r^2}}$$

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