# What is the area of the circle?

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. • It is a disc, not a circle :)
– Vim
Jan 24, 2016 at 16:04

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. : • Nice diagrams. How did you draw them?
– Karl
Jan 24, 2016 at 16:29
• @Karl I'm guessing paint. Jan 25, 2016 at 2:13
• @ThisIsNotAnId You guessed it right! Jan 25, 2016 at 5:02 Let $$r$$ be the radius of the circle with center $$O$$. Join the vertex of contact $$P$$ of small rectangle to the center $$O$$. Draw two perpendicular lines $$PQ$$ & $$OQ$$ to get right $$\Delta PQO$$ (as shown in above figure). Using Pythagorean theorem in right $$\Delta PQO$$ as follows $$r^2=(r-3)^2+(r-4)^2$$ $$r^2-14r+25=0$$ 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. For $$r=7-2\sqrt 6$$ to be valid, the rectangle has to be touching the circle internally. But in the given figure, the case is the exact opposite hence $$r=7+2\sqrt 6$$ is acceptable.

Substituting the value of $$r$$, calculate area of circle $$=\color{red}{\pi (7+2\sqrt6)^2\approx 444.8046352 }$$

• 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. Jan 25, 2016 at 9:05

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 $r-4$.

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 $R\ge 4$.