# Analytic Geometry Question (equation of a circle)

## 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).

## 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:

eumyang
Homework Helper
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. So the slope of the perpendicular will instead be 3/5, and the equation of the line will be... ?

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.

eumyang
Homework Helper
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.

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.

ehild
Homework Helper
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

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