Theorem 3.3: The circle whose equation is
and the line
meet at the points where
Proof: If we substitute the value for y in the equation of the line in for the y in the equation of the circle we get
combine like terms, and transpose all the terms to the left, leaving a 0 on the right.
We can now use the quadratic formula to solve for x. If ax2 + bx + c = 0, the reader can verify that
will simplify to the familiar quadratic formula. However, it will prove simpler in our case.
Let us simplify the radicand:
Remove parentheses and combine like terms
= (m2 + 1)r2 - (y02 + m2x02+ b2 - 2mx0y0 + 2mbx0 - 2by0)
We substitute this into the radical, and we get the following solution for x
To get the y coordinate of the points of intersection, we need only substitute this solution for x into either of the original equations. It would be easier to substitute it into the linear equation.
We could combine the radical free terms by finding common denominators, etc., but that will not be necessary. We know from Theorem 1.10 that the quantity inside the parentheses in the first term is the x-coordinate of the foot of (x0, y0) in the line y = mx + b. If we substitute the x-coordinate of a point on the line into the equation, we get the y-coordinate of the point. We know from Theorem 1.10 that the y-coordinate of the foot of (x0, y0) in the line y = mx + b is
We substitute this in and get
While these forms of the formulas will prove to be more useful for actually computing the coordinates of the points where a line meets a circle, there is another form which will prove most useful.
Consider the radical term in the formula for the x-coordinates.
Factor (m2 + 1) from out of both terms.
We can further irrationalize the denominator as
This leads us to the formulas
next theorem (3.4)