# Coordinate geometry of the circle question

## Homework Statement

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.

## Homework Equations

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

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

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

## 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!

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HallsofIvy
Homework Helper
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.

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

this has complex roots as well....

Dick
Homework Helper
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).

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?

Dick
Homework Helper
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?
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).