# Radius of the circumscribed circle

We have a triangle with $a=b$. I want to calculate the radius of the circumscribed circle in respect to $a$ and $c$.

We have that the center is the point where the perpendicular bisectors intersect, and the radius is the length to any of the three vertices.



From the sinus law we have that $$\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R$$ where $R$ is the radius of the circumscribed circle.

So, we have that $2r=\frac{a}{\sin\alpha}$. Do we have to write $\sin\alpha$ in respect to $c$ ?

• You can write $\cos \alpha$ in terms of $a$ and $c$. Then you can derive $\sin \alpha$ from there. – Paul Aljabar Jul 19 '17 at 17:30
• Do you mean by the cosine law? @PaulAljabar – Mary Star Jul 19 '17 at 17:34

## 2 Answers

Do you fear drawing diagrams? They might help. Quite a lot. By considering the midpoint of $AB$ and applying the Pythagoran theorem we have $$R+\sqrt{R^2-\frac{c^2}{4}} = \sqrt{a^2-\frac{c^2}{4}}$$ That is simple to turn into a quadratic equation, leading to $R=\frac{a^2}{\sqrt{4a^2-c^2}}$.

• I got stuck right now... At which triangle do we apply the pythagorean? Let $O$ be the center and M the midpoint of $AB$. Do we apply the pythagorean theorem at $AOM$ ? But how do we calculate $OM$ ? – Mary Star Jul 19 '17 at 17:41
• @MaryStar: consider the altitude from $C$. Its length can be computed by applying the Pythagorean theorem to half of $ABC$, leading to the RHS. On the other hand, its length is given by $R+OM$, and $OM$ can be computed by applying the Pythagorean theorem to $OMA$: $$OM=\sqrt{AO^2-AM^2}=\sqrt{R^2-\frac{c^2}{4}}.$$ – Jack D'Aurizio Jul 19 '17 at 17:43
• I understand!! Thank you so much!! :-) – Mary Star Jul 19 '17 at 18:17

Using the symmetry of the isosceles triangle, you have $$\cos\frac{C}{2}=\frac{a}{2R}\implies\cos C=\frac{a^2}{2R^2}-1$$

Also, from the sine rule, you have $\sin C=\frac{c}{2R}$, so combining these gives $$\frac{c^2}{4R^2}+\left(\frac{a^2}{2R^2}-1\right)^2=1$$

This can be rearranged to give $$R^2=\frac{a^4}{4a^2-c^2}$$ and hence $R$ in terms of $a$ and $c$

• How do we get that $\cos\frac{C}{2}=\frac{a}{2R}$ ? By the cosine law? – Mary Star Jul 19 '17 at 17:43
• @MaryStar by considering the right angled triangle formed by the midpoint of $AC$, the centre of the circle and $C$ – David Quinn Jul 19 '17 at 18:35
• Ah ok!! Thank you very much!! :-) – Mary Star Jul 19 '17 at 23:23