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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 $c$ ?

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    $\begingroup$ You can write $\cos \alpha$ in terms of $a$ and $c$. Then you can derive $\sin \alpha$ from there. $\endgroup$ – Paul Aljabar Jul 19 '17 at 17:30
  • $\begingroup$ Do you mean by the cosine law? @PaulAljabar $\endgroup$ – Mary Star Jul 19 '17 at 17:34
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Do you fear drawing diagrams? They might help. Quite a lot. enter image description here

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

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  • $\begingroup$ 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 $OM$ ? $\endgroup$ – Mary Star Jul 19 '17 at 17:41
  • $\begingroup$ @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 $OMA$: $$OM=\sqrt{AO^2-AM^2}=\sqrt{R^2-\frac{c^2}{4}}.$$ $\endgroup$ – Jack D'Aurizio Jul 19 '17 at 17:43
  • $\begingroup$ I understand!! Thank you so much!! :-) $\endgroup$ – Mary Star Jul 19 '17 at 18:17
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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 $c$

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

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