# Basic question: Is it possible for the equation of a circle to have an xy term?

1. Jun 4, 2009

### seeker101

Can this be an equation of a circle $$x^2+y^2-2hx-2ky+2mxy+r=0$$

2. Jun 4, 2009

### slider142

No. The presence of xy denotes a term of the form (Ax + By)2, which does not form any part of the general equation of a circle. It may, however, be a conic of some sort that is simply rotated. You can get cross terms by applying a rotation change of coordinates (x, y) -> (x', y') to the general equation of a conic, x' = Cx - Sy, y' = Sx + Cy, where S= sin(t) and C = cos(t) where t is the angle of rotation. In the case of a circle, the cross terms cancel.
You can find the general equation of a conic section here.

3. Jun 5, 2009

### HallsofIvy

Staff Emeritus
The presence of "xy" indicates rotation of the axes. And a circle looks exactly the same no matter how it is rotated!

4. Jun 5, 2009

### AUMathTutor

Perhaps, if m = 0.

5. Jun 5, 2009

### quZz

I guess it can when m = 0, r < k2 + h2

6. Jun 5, 2009

### HallsofIvy

Staff Emeritus
The question in the title was "Is it possible for the equation of a circle to have an xy term" to which the correct answer is "No".

The question in the text was "Can this be the equation of a circle $x^2+ y^3- 2hx- 2kx+ 2mxy= r$" to which the answer is "Yes". It is possible with m= 0 (so the equation does not have a an xy term) and $h^2+ y^2< r$, NOT $r< h^2+ k^2$. If m= 0 and $h^2+ k^2= r$ the graph is the single point (h, k). If m= 0 $h^2+ k^2> r$ there are no (x, y) points satifying the equation.