# Analytic Geometry Question (equation of a circle)

1. Jul 7, 2012

### vodkasoup

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

3. 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.

Last edited: Jul 7, 2012
2. Jul 7, 2012

### eumyang

The slope is -5/3, not 5/3. So the slope of the perpendicular will instead be 3/5, and the equation of the line will be... ?

3. Jul 7, 2012

### vodkasoup

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. Jul 7, 2012

### eumyang

I got 3x - 5y + 16 = 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. Jul 8, 2012

### ehild

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. Jul 9, 2012

### vodkasoup

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