Coordinate geometry of the circle question

[Ed Aboud]

Homework Statement

Fairly straight forward question but I just can't see what's 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}

Where r is the radius.

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!

 

[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.

 

[Ed Aboud]

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

this has complex roots as well...

 

[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).

 

[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?

 

[Dick]

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).