# Coordinate geometry of the circle question

1. Feb 21, 2009

### Ed Aboud

1. The problem statement, all variables and given/known data
Fairly straight forward question but I just can't see whats wrong.

A circle passes through the point (2,-4) and touches both the x-axis and the y-axis. Find the equations of the two circles which satisfy these conditions.

2. Relevant equations

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

with a centre point $c (-g,-f)$

$$r = \sqrt{g^2 + f^2 - c}$$

3. The attempt at a solution

After drawing a diagram I concluded that r=g and r=f therefore g=f.

$$g = \sqrt{g^2 + f^2 - c}$$

$$g^2 = g^2 + f^2 - c$$

$$f^2 = c$$

Since the point (2,-4) is on the circle it will satisfy :

$x^2 + y^2 + 2gx + 2fy + c = 0$

$$(2)^2 + (-4)^2 + 2g(2) + 2f(-4) + c = 0$$

$$4 + 16 + 4g -8f + c = 0$$

$$20 + 4g -8f + c = 0$$

But $f^2 = c$ and $f = g$

So

$$20 + 4f - 8f + f^2 = 0$$
$$20 -4f + f^2 = 0$$
$$f^2 -4f +20 = 0$$

This quadratic only has complex roots.

Thanks for any help!

2. Feb 21, 2009

### HallsofIvy

If the center is $(x_0,y_0)$, that is, the equation of the circle is $(x-x_0)^2+ (y-y_0)^2= R^2$ and it "touches" the axes- by which I assume you mean it is tangent to them, then $(x_0,0)$ and $(0,y_0)$ are the points at which the circle is tangent to the axes. Putting $x= x_0$, y= 0 into the equation, $(-y_0)^2= R^2$ and putting $y= y_0$, x= 0, $(-x_0)^2= R^2$ so we must have $x_0= y_0= R$ so we can write the equation as $(x- R)^2+ (y- R)^2= R^2$. Adding the condition that (2,-4) lies on that circle gives $(2-R)^2+ (-4-R)^2= R^2$ which you can solve for R.

3. Feb 23, 2009

### Ed Aboud

Yeah so you get $$R^2 + 4R + 20 = 0$$

this has complex roots as well....

4. Feb 23, 2009

### Dick

The center of the circle isn't at (R,R). That's in the wrong quadrant. You have to pick y0 to be the negative root of y0^2=R^2. Put it at (R,-R).

5. Feb 23, 2009

### Ed Aboud

Ok cool thanks so it is

$$R^2 -12R +20 = 0$$

$$R=2$$ and $$R=10$$

Therefore the equations of the circles are

$$(x- R)^2+ (y+ R)^2= R^2$$

$$(x- 2)^2+ (y+ 2)^2= R^2$$

and

$$(x- 10)^2+ (y+ 10)^2= R^2$$

Just out of curiosity what was wrong with my original attempt?

6. Feb 23, 2009

### Dick

Exactly the same thing. You said f=g=r. But that's not true. |f|=|g|=r is what is true. You have to figure out the signs based on the location of (2,-4).