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In the following diagram, $AB = 4$ and $AC = 3$. What is the area of the circle? I can't find any way to solve this.

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It is a disc, not a circle :) – Vim Jan 24 at 16:04
up vote 10 down vote accepted

let $r$ be the radius of the circle.

Join the vertex of contact of small rectangle to the center of the circle & drop two perpendiculars from the vertex of contact to vertical & horizontal radial lines.

Consider a right triangle with hypotenuse $r$ & legs $r-3$ & $r-4$

then using Pythagorean theorem $$r^2=(r-3)^2+(r-4)^2$$ $$r^2=r^2-6r+9+r^2-8r+16$$ $$r^2-14r+25=0$$ Now, solve the above quadratic equation to find the values of $r$ as follows $$r=\frac{-(-14)\pm\sqrt{(-14)^2-4(1)(25)}}{2(1)}=7\pm2\sqrt{6}$$ There are two cases, but for $r=7-2\sqrt 6$, the rectangle is touching the circle internally but in the given figure rectangle is touching the circle externally hence $r=7+2\sqrt 6$ is acceptable. Hence, substituting the value of $r$, calculate area of circle $$=\color{red}{\pi (7+2\sqrt6)^2\approx 444.8046352 }$$

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I think the emergence of the second (non-applicable) solution here is the best part. Finding things like this, that you weren't initially looking for, is one of the beauties of mathematics for me. – Svj0hn Jan 25 at 9:05

let $r$ be radius. Then equation of circle with center (0,0) and radius r is $x^2 + y^2 = r^2$

$-r+3$ and $r-4$ satisfy this equation. Therefore place them and get the value of $r$.

$$ (-r+3)^2 + (r-4)^2 = r^2$$

$$r^2 -14r +25 = 0$$

On solving, there are two possibilities: one is about 2.1, other is 11.9. We logically reject the first value, as the rectangle formed thus would touch it internally. :

enter image description here

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Nice diagrams. How did you draw them? – Karl Jan 24 at 16:29
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@Karl I'm guessing paint. – ThisIsNotAnId Jan 25 at 2:13
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@ThisIsNotAnId You guessed it right! – Max Payne Jan 25 at 5:02

A start: Let the radius be $r$ and the centre of the circle be $O$. Let the lower right-hand corner of the little rectangle be $P$. Draw the line $OP$. Note $OP$ is the hypotenuse of a right triangle with legs $r-3$ and $r-4$.

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Hint:

Let $M$ be the fourth vertex of the rectangle (on the circle), and denote $x$ its polar angle, w.r.t. the polar axis through the centre of the circle. You can express $\sin x$ and $\cos x$ as functions of the radius $R$ of the circle. Then write down Pythagoras' identity $\;\sin^2x+\cos^2x=1\;$ to obtain a quadratic equation for $R$. Don't forget the solution, if any, is subject to the condition $R\ge 4$.

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