Find the excentre of the triangle formed by the points (0,3), (4,0), and (0,0) which is opposite to (0,0) Even though, my teacher had informed me that this isn't a highschool question, I wanted to know how you can calculate the excentre? 
PS: I really love Mathematics! 
 A: The excircle in question is tangent to the coordinate axes. Thus the excentre has the form $(r,r)$ where $r$ is the exradius.
To find $r$ consider the following diagram (top-right vertex is excentre, lower-left vertex is origin):

By similar triangles we get
$$\frac{3-\frac15r}{\frac25r}=\frac34\implies r=6$$
and the excentre is $(6,6)$.
A: Let $A(0,0)$, $B(0,3)$, $C(4,0)$ be the given points.
The excenter w.r.t. $A$ is $I_A(6,6)$ in the picture:

First of all, it is on the angle bisector of the angle $\widehat{BAC}$, which is the first bisector in the coordinate system. So it is a point of the shape $(s,s)$. Then the distance from $(s,s)$ to the line $BC$ given by the equation $BC:\ 3x+4y-12=0$, or better in normed form
$$
BC\ :\ \frac 35 x+\frac 45 y -\frac{12}5 =0\ ,
$$
is obtained by plugging in $(s,s)$ in the above "normed equation". We get $(7s-12)/5$. (And have to take absolute value of it.) The distance from $(s,s)$ to the axes (which are corresponding to the other two sides) is $s$. So we get the equation $s=(7s-12)/5$. The solution is $s=6$.

Comment: An other way to get the touching point $P$ of the excircle with the hypotenuse $BC$ is to compute the half-perimeter $p$ of the triangle $ABC$ with sides $3,4,5$, it is $p=(3+4+5)/2=6$. Now the point $P$ is on $BC$ at distances $p-b=6-4=2$, and $p-c=6-3=3$ from $C$, respectively $B$. (Together we get $2+3=5=BC$.) 
A: I’m not sure if I got your question: you’re looking for the coordinates of the center of the circle tangent to the side opposite to vertex (0,0) and the extensions of the other two sides?
If answer is yes, then the problem is high school level of analytic geometry.
I hope my writing is not impossible

