Consider a circle $ C $ with radius $ r $ and a point $P $ outside the circle. Construct two tangents from $P$ to the circle, touching the circle at points $ A $ and $ B $. Let $ O $ be the center of the circle.
Question: If $ OP = 2r $ , find the exact length of the tangent segment $ PA $ (from point $ P $ to point $A $.
My approach: Now, let me give this problem a try.
Given that the area of triangle OAB is $3r^2$ and OP is twice the length of the radius $2r$, I need to determine the length of the tangent segment PA.
First, I denoted the length of PA as $x$. Since triangle $OAB$ is isosceles, we know that $OA = OB$.
Using the Pythagorean theorem, I expressed OB in terms of r: $OB^2$ = $OP^2$ - $BP^2$ $OB^2$ = $2r^2$ - $r^2$ $OB^2$ = $4r^2$ - $r^2$ $OB^2$ = $3r^2$
Next, I find the area of triangle OAB: $[Area_{OAB} = \frac{1}{2} \times PA \times OB]$ $[3r^2 = \frac{1}{2} \times x \times \sqrt{3r^2}]$
Solving for $(x)$: $[6r^2 = x \times r \sqrt{3}]$ $[x = \frac{6r^2}{r \sqrt{3}}]$ $[x = \frac{6r}{\sqrt{3}}]$ after rationalizing $[x = 2r \sqrt{3}]$
So, according to my calculations, the length of the tangent segment PA is $2r \sqrt{3}$. This seems like a plausible solution, but but it seems that the correct answer to this question is $3r$. Where did I go wrong and how can we arrive at the correct answer
This is a rather odd question. Since we know that $PA$ is tangent to the circle we immediately know that $OA$ is a radius. Of course, this means it has length $r$, and also that $PAO$ is a right-angle. Additionally, you state that $OP$ has length $2r$. This trivially allows the Pythagorean theorem to be used to solve for $x$.
$$ \begin{aligned} r^2 + x^2 & = (2r)^2 \\ r^2 + x^2 & = 4r^2 \\ x^2 & = 3r^2 \\ x & = \sqrt{3} r \end{aligned} $$
The area information is not used. I didn't check if it is consistent with everything else. It may just be a misdirect, or the problem may be overspecified in some incorrect way, but you can check this easily enough yourself.