# Find the center of a circle on the x-axis with only two points, no radius/angle given

Find the center $C$ on the x-axis of the circle containing $(15,-2)$ and $(7,10)$

I can't seem to find a formula to help me solve this problem without needing the radius or the angle between the the two points.

• Hint: $CX = CY$. – aschepler Aug 8 '14 at 16:44

The ordinate of any point on the x-axis is zero, so, any point on the x-axis can be written as $(a,0)$

So, we have $$\sqrt{(a-15)^2+(0+2)^2}=\sqrt{(a-7)^2+(0-10)^2}$$

Find $a$

So, the equation of the circle $$(x-a)^2+(y-0)^2=(a-15)^2+(0+2)^2$$

Since $C$ is the same distance from (15,−2) and (7,10), it must be on the perpendicular bisector of those two points. So find the midpoint and slope of the segment between those points, find the slope of the perpendicular bisector (the negative reciprocal of the segment), and use the point and slope to get the equation of the perpendicular bisector. In that equation, set $y=0$ (the equation of the x-axis) and you find the x-coordinate of $C$. Then you know where $C$ is.

Let us know if you need help with any of those steps.

We can avoid the squaring and taking roots if we use the general equation for the circle, viz (xsqd + ysqd + 2gx + 2fy +c =0 whose centre is given by (-g, -f).Here, f = 0 since centre is on x-axis.Solving the simultaneous equations we gwt the centre as (5, 0).

Hint:

Use the fact that it lies on x-axis, i.e. $C\equiv(\alpha,0)$ Secondly distance to both points must be same and equal to radius