Why does the equation of a circle have to have the same $x^2,y^2$ coefficients? In one of my geometry texts, it tells me they should be the same but not why. I am unsatisfied with this.
Suppose that:
$$ax^2+by^2 + cx + dy + f = 0 \text{ such that } a \neq b$$
is the equation of some circle. Upon completing the square and rearranging, I obtain
$$a\left(x+\frac{c}{2a}\right)^2 + b\left(y+\frac{d}{2b}\right)^2 = \frac{c^2}{4a} + \frac{d^2}{4b} - f$$
I know that a circle is defined as the set of points a fixed distance from a fixed point.
How can i arrive at a satisfying contradiction? At the moment I just can't see it. 

****According to the definition of a circle , it's a locus of a points from a fixed point at a constant distant ****
Now take a fixed point C( h,k)which we call it Center of circle and constant distance is 'r' which is it's radius 
Let $P(x,y)$ be any point in locus 
By the definition we have 
$PA=r$
The distance between Center and a any locus point is constant 'r'
$ \sqrt{(x-h)^2+(y-k)^2}=r$
Squaring on both sides 
$ (x-h)^2+(y-k)^2=r^2$
$=>x^2+h^2-2xh+y^2+k^2-2yk=r^2$
$=>x^2+y^2-2hx-2ky+h^2+k^2-r^2=0$
From
 this equation we have the following points which are characters of circle 


*

*circle is a second degree equation 

*$coefficient of x^2=coefficient y^2$

*$coefficient of xy=0$

*$ radius of circle r>=0$



Now compare your circle equation with this properties implies  a=b

