Theorem 3.13: Given two circles whose equations are
and
their points of intersection are given by
and
where
Proof: The points of intersection lie on a line whose equation is given from Theorem 3.10.
We will consider two cases. The first is where the centers are not horizontal, and the second is where they are.
Case 1: The centers are not horizontal. We can use the simpler formula for the points where a line intersects a circle from Theorem 3.3
and
where by Theorem 3.10,
and
We know what will happen if we substitute these values for m and b into the rational term of the formula for the x coordinate because we did that in the proof of Theorem 3.12, and we got the rational terms in the expression for x in this result. So we need only verify that the radical term gives us what we want. Let us first work with the expression under the radical.
This becomes
Find common denominators.
While we remove the parentheses in the second term, let's add the numerators inside the first parentheses.
Combine like terms in the numerator of the second fraction.
This can be rewritten as
Let
be the distance between the centers. Our expression can then be written as
We we find common denominators for these two fractions.
We can now factor the top as a difference of squares.
Remove the inside parentheses, and rearrange the terms.
or
Both factors on top are differences of squares, and will factor.
We rearrange the factors
and we find that we have two sum time difference multiplications which we can multiply as
We put this back into the radical and get
Rationalize the denominator under the radical and simplify the denominator.
If we simplify the compound fraction we get
and we get the radical term in the expression for x in the statement. Since we will be taking both the positive and negative square root, we can drop the absolute value signs around the difference in the y-coordinates.
To get the y-coordinate of the points of intersection we can substitute this value of x into the equation from Theorem 3.10. This would give us
where
and
We could write this as
We know how the first two terms will simplify. We are substituting the x-coordinate of the point where the line joining the centers of the circles meets the line containing the points of intersection into the equation of the line containing the points of intersection, so we will get the y coordinate of that point which we know from Theorem 3.12 to be
In the radical term, the difference in the y-coordinates cancels out and we are left with
and we get the desired result.
Case 2: y1 = y2. In this case the line which contains the points of intersection has the equation
by Theorem 3.12.
In the expression for the x-coordinates of the points of intersection of the two circles in the statement of this result under the condition that y1 = y2 will both be the same and will simplify to the right side of this equation.
The y-coordinates of the points of intersection between the line and the circle are
by Theorem 3.2. This needs to be simplified. Find common denominators inside the parentheses under the radical.
Remove the inside parentheses.
While we combine like terms inside the parentheses, let's also find common denominators for the two terms under the radical.
We can now rationalize the denominator and make an adjustment in the second term on the top under the radical.
In this case, |x1 - x2| = d, so we can simplify this expression.
The numerator under the radical is the same one that we had in the other case and it will similarly factor to
and in the case where y1 = y2, this is the formula to which the formula for the y-coordinates of the points where the circles intersect will simplify.
Next section: 4. Translations and Reflections