Linear Intersections of Circles (Algebraic Method)

Brian Swanagan

Suppose that you have two circles with equal radius intersecting each other at two points.  Define a coordinate plane such that (0,0) are the coordinates of one of the points of intersection with the other point lying along the y-axis and (1,p) are the coordinates of one of the centers of the circles.

Then the other center has coordinates of the other center are (-1,p) because it both intersection points lie on the y-axis and the centers are equidistant from those two points and so fall on the perpendicular bisector to the segment connecting the intersections and further has the same distance away from those intersections as the first center.

The other intersection then has coordinates (0,2p) because the line through the centers is y=p and so (0,p) is the midpoint between them requiring the point to have the stated coordinates.

Suppose we have a line that passes through (0,0).  Then, it has the equation y=mx for some m.

The radii of the two circles are (p2+1).5.  And so, the equations of the two circles are (x-1)2 + (y-p)2 = (p2+1) and (x+1)2 + (y-p)2 = (p2+1).

We would like to determine the intersection of m with the two circles.  First, letŐs use substitution to find the intersection point of the line with our first circle equation.

So,

(x-1)2 + (mx-p)2 = (p2+1).

x2 – 2x + 1 + m2x2 – 2mpx + p2 = p2 + 1

(m2+1)x2 – 2(1+mp)x = 0

x[(m2+1)x – 2(1+mp)] = 0

This implies that x = 0 or x = 2(1+mp)/(m2+1) and that y = 0 or 2m(1+mp)/(m2+1) respectively.

Now, letŐs use substitution to find the intersection point of the line with our second circle equation.

So,

(x+1)2 + (mx-p)2 = (p2+1).

x2 + 2x + 1 + m2x2 – 2mpx + p2 = p2 + 1

(m2+1)x2 – 2(mp-1)x = 0

x[(m2+1)x – 2(mp-1)] = 0

This implies that x = 0 or x = 2(mp-1)/(m2+1) and that y = 0 or 2m(mp-1)/(m2+1) respectively.

In both these cases, we know that (0,0) is the point of intersection of the 2 points and the line and the other two points are of the line and with only one of two circles.

We would like to know if the two points other than the intersection (0,0) on this line are equidistant from the other intersection point located at (0, 2p).

We will find the square of the distance for now since that will be equal when the distances are equal as well to make it a bit easier to work with.

[(2(mp-1)/(m2+1) – 0]^2 + [2m(mp-1)/(m2+1) – 2p]2

= 4(mp-1)2/(m2+1)2 + 4m2(mp-1)2/(m2+1)2 – 8mp(mp-1)/(m2+1) + 4p2

= [4(mp-1)2 + 4m2(mp-1)2 – 8mp(mp-1)(m2+1)]/(m2+1)2 + 4p2

= [4m2p2 – 8mp + 4 + 4m4p2 – 8m3p + 4m2 – 8m4p2 + 8m3p – 8m2p2 + 8mp]/(m2 + 1) + 4p2

= [-4m4p2 – 4m2p2 + 4m2 +4]/(m2+1)2 + 4p2

= 4(1-m2p2)(m2+1)/(m2+1)2 + 4p2

= 4(1-m2p2)/(m2+1) + 4p2

= [4 – 4m2p2 + 4m2p2 + 4p2]/(m2+1)

= 4(p2+1)/(m2+1).

And, for the other point:

[2(1+mp)/(m2+1) – 0]2 + [2m(1+mp)/(m2+1) – 2p]2

= 4(mp+1)2/(m2+1)2 + 4m2(mp+1)2/(m2+1)2 – 8mp(mp+1)/(m2+1) + 4p2

= [4(mp+1)2 + 4m2(mp+1)2 – 8mp(mp+1)(m2+1)]/(m2+1) + 4p2

= 4[m2p2 + 2mp + 1 + m4p2 + 2m3p + m2 – 2m4p2 – 2m3p – 2m2p2 – 2mp]/(m2+1) + 4p2

= 4[-m4p2 – m2p2 + m2 + 1]/(m2+1) + 4p2

= 4(1-m2p2)(m2+1)/(m2+1)2 + 4p2

= 4(1-m2p2)/(m2+1) + 4p2

= (4 – 4m2p2 + 4m2p2 + 4p2)/(m2+1)

= 4(p2+1)/(m2+1).

We can see that the square of the distance from the two points is the same to the other intersection and so they lie on a circle with radius 4(p2+1)/(m2+1) with center (0,2p).

Any circle that centered on another circle will intersect with that circle if and only if itŐs radius r>0 is no more than the diameter of that circle.

So, our new circle centered at (0, 2p) intersects the original two circles if 0<r<2(p2+1).5 or equivalently 0 < r2 < 4(p2+1).

But notice the range of possible values for our distances from the intersection, (0,2p), from the points created by our line with each circle.

m2 > 0 so m2 + 1 > 1 with no upper bound on each so 0 < 1/(m2+1)< 1 and finally 0 < 4(p2+1)/(m2+1) < 4(p2+1) and r2 = 4(p2+1)/(m2+1) when we create a circle through these two points so we can see that r2 indeed has the necessary range.  That implies that we ought to be able to determine m in terms of r so that we can find a line that intersects the two circles at 2 points and (0,0) such that those first 2 points are a distance of r away from (0,2p).

We shall solve quickly for m

4(p2+1)/(m2+1) = r2

4(p2+1)/r2 = m2+1

4(p2+1)/r2 – 1 = m2

+/- [4(p2+1)/r2 – 1].5 = m

So, we find two values for our lineŐs slope (the line being y=mx) that passes through (0,0) that contains 2 more points intersecting the two circles that are a distance of r away from (0,2p) which means they lie on a circle with radius r constructed with a center at that point that intersects those the two original circles.  Then, for any circle centered at (0,2p) that intersects those two circles, one intersection from each of those circles and (0,0) are colinear.

Why do we get two values?  Notice that the two lines are symmetric around x=0 which is reasonable because we constructed our coordinate system to be symmetric to the y-axis which is x=0 and so the intersects created by the circle centered at (0,2p) are symmetric to x=0 giving us two possible pairs of intersections with the two circles that are also colinear with (0,0) except when m=0 in which case r = 2(p2+1) so that the circle centered at (0, 2p) only intersects each of the original circles in one place.  Then, we only have one possible line y=0 that passes through those and (0,0).