Analytic Geometry Question (equation of a circle)

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  • #1
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Homework Statement



Find the equation of the circle whose centre lies on the x axis and which passes through points A (6,0) and B (0,10).

Homework Equations





The Attempt at a Solution



I drew a diagram of the circle and determined that the line AB has gradient 5/3. Its perpendicular bisector should pass through the centre of the circle. This perpendicular line has a gradient of -3/5 and, I think, an equation of 3x+5y-34=0.

The centre lies on the x-axis, so the y-coordinate at the centre must be 0. Hence I can use the equation of the perpendicular line to solve for x at point (x,0). I get 34/4 for x here, giving a centre at (34/3 ,0).

I am not sure where to go now, or if the above is even correct. Any help greatly appreciated.
 
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Answers and Replies

  • #2
eumyang
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I drew a diagram of the circle and determined that the line AB has gradient 5/3.
The slope is -5/3, not 5/3. :wink: So the slope of the perpendicular will instead be 3/5, and the equation of the line will be... ?
 
  • #3
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Whoops...I'll blame that one on lack of sleep! ;)

So I'm on the right track? The equation of the perpendicular line is now (I hope) 3x-5y-16=0.
 
  • #4
eumyang
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Whoops...I'll blame that one on lack of sleep! ;)

So I'm on the right track? The equation of the perpendicular line is now (I hope) 3x-5y-16=0.
I got 3x - 5y + 16 = 0. :wink:

The centre lies on the x-axis, so the y-coordinate at the centre must be 0. Hence I can use the equation of the perpendicular line to solve for x at point (x,0).

After you do this, find the distance from the center to either A or B. That will give you the radius. Once you have the center and radius, you can write the equation of the circle.
 
  • #5
ehild
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You can find the centre of the circle without much geometry, just using the fact that the given points are at equal distance R from it. If the coordinates of the centre are (a,0), a2+102=(6-a)2=R2.

ehild
 
  • #6
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Thanks for your replies ehild and eumyang, you were a great help.
 

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