Radius of the circumscribed circle

Question

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.

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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 $c$ ?

Answer 1

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 $R=\frac{a^2}{\sqrt{4a^2-c^2}}$.

Answer 2

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 $c$