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Coordinate geometry of the circle question

  • Thread starter Ed Aboud
  • Start date
  • #1
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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



[tex] x^2 + y^2 + 2gx + 2fy + c = 0[/tex]

with a centre point [itex] c (-g,-f) [/itex]

[tex] r = \sqrt{g^2 + f^2 - c} [/tex]

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.

[tex] g = \sqrt{g^2 + f^2 - c} [/tex]

[tex] g^2 = g^2 + f^2 - c [/tex]

[tex] f^2 = c [/tex]

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

[itex] x^2 + y^2 + 2gx + 2fy + c = 0[/itex]

[tex] (2)^2 + (-4)^2 + 2g(2) + 2f(-4) + c = 0 [/tex]

[tex] 4 + 16 + 4g -8f + c = 0 [/tex]

[tex] 20 + 4g -8f + c = 0 [/tex]

But [itex] f^2 = c [/itex] and [itex] f = g [/itex]

So

[tex] 20 + 4f - 8f + f^2 = 0[/tex]
[tex] 20 -4f + f^2 = 0 [/tex]
[tex] f^2 -4f +20 = 0 [/tex]

This quadratic only has complex roots.

Thanks for any help!
 

Answers and Replies

  • #2
HallsofIvy
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Homework Helper
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If the center is [itex](x_0,y_0)[/itex], that is, the equation of the circle is [itex](x-x_0)^2+ (y-y_0)^2= R^2[/itex] and it "touches" the axes- by which I assume you mean it is tangent to them, then [itex](x_0,0)[/itex] and [itex](0,y_0)[/itex] are the points at which the circle is tangent to the axes. Putting [itex]x= x_0[/itex], y= 0 into the equation, [itex](-y_0)^2= R^2[/itex] and putting [itex]y= y_0[/itex], x= 0, [itex](-x_0)^2= R^2[/itex] so we must have [itex]x_0= y_0= R[/itex] so we can write the equation as [itex](x- R)^2+ (y- R)^2= R^2[/itex]. Adding the condition that (2,-4) lies on that circle gives [itex](2-R)^2+ (-4-R)^2= R^2[/itex] which you can solve for R.
 
  • #3
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Yeah so you get [tex] R^2 + 4R + 20 = 0[/tex]

this has complex roots as well....
 
  • #4
Dick
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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
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Ok cool thanks so it is

[tex] R^2 -12R +20 = 0 [/tex]

[tex] R=2 [/tex] and [tex] R=10 [/tex]

Therefore the equations of the circles are

[tex] (x- R)^2+ (y+ R)^2= R^2 [/tex]

[tex] (x- 2)^2+ (y+ 2)^2= R^2 [/tex]

and

[tex] (x- 10)^2+ (y+ 10)^2= R^2 [/tex]

Just out of curiosity what was wrong with my original attempt?
 
  • #6
Dick
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Homework Helper
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Ok cool thanks so it is

[tex] R^2 -12R +20 = 0 [/tex]

[tex] R=2 [/tex] and [tex] R=10 [/tex]

Therefore the equations of the circles are

[tex] (x- R)^2+ (y+ R)^2= R^2 [/tex]

[tex] (x- 2)^2+ (y+ 2)^2= R^2 [/tex]

and

[tex] (x- 10)^2+ (y+ 10)^2= R^2 [/tex]

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).
 

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