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Given triangle $\triangle ABC$ , with $AB=AC$ and 2 times(length of inradius) which is equal to the length of exradius of excirle opposite vertex $A$. Setup the coordinate system in the plane of triangle $\triangle ABC$ as follow: Origin at vertex $B$, positive X-axis along $BC$ and positive Y-axis on the $A$ side of $BC$. Find the equations of the incircle , excircles and circumcircle.

Also this question is related to the incenter excenter configuration of triangle.I would like to get some help to solve the above question.

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it was not that i did not accept the answers but i was really not knowing what we should do to show that i have accepted the answer....really sorry for that – mgh Jun 12 '12 at 13:48
You can find the equation of incircle here. – Gigili Jun 12 '12 at 16:57

Hint: Here is a diagram of the construction described above.

$\hspace{4.5cm}$enter image description here

There is no absolute scale given in the problem, so lets set the inradius of $\triangle ABC$ to be $1$. Since the radius of the excircle opposite $A$ is twice that of the incircle, it is $2$. By similar triangles, we have $|AE|=2|AD|$. Furthermore, $|DE|=3$, so $|AD|=3$ and $|AE|=6$.

Everything else should be calculable using similar triangles.

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In the coordinate system $(B,\vec{BC},\vec{BC'})$ we have $B(0,0)$, $C(1,0)$ and $A\left(\frac 12,a\right)$ where $a=AH$ : height from vertex $A$. You have to find $a$ using hypothesis and the rest must be easy.

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:How did you get the coordinates of vertex B and vertex C? – mgh Jun 12 '12 at 13:18
"the rest must be easy", but it is not. – Gigili Jun 12 '12 at 15:48
: *How did you get the coordinates of vertex A and vertex C? – mgh Jun 12 '12 at 16:07
@ meg_1977 and Gigili : Since geometric properties do not change by changing the unit of reference, we can choose BC for the unit, then coordinates of $C$ are $1$ and $0$.The abscissa of $A$ is $\frac 12$ and it's ordinate is $a=AH$ where $H$ is the middle of $[BC]$ who can be calculated by using radius hypothésis. – Mohamed Jun 12 '12 at 22:15

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