Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
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 $$=\color{red}{\pi (7+2\sqrt6)^2\approx 444.8046352 }$$

share|cite|improve this answer
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

share|cite|improve this answer
Nice diagrams. How did you draw them? – Karl Jan 24 at 16:29
@Karl I'm guessing paint. – ThisIsNotAnId Jan 25 at 2:13
@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 $r-4$.

share|cite|improve this answer


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 $R\ge 4$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.