**Theorem
3.3**: The circle whose
equation is

and the line

meet at the points where

and

or

and

**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

Remove parentheses,

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 *ax*^{2} + *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

= (*m*^{2} + 1)*r*^{2} - (*y*_{0}^{2} + *m*^{2}*x*_{0}^{2}+ *b*^{2} - 2*mx*_{0}*y*_{0} + 2*mbx*_{0} - 2*by*_{0})

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.

or

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 (*x*_{0}, *y*_{0}) 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 (*x*_{0},
*y*_{0}) 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 (*m*^{2} + 1) from out of both terms.

We can further irrationalize the denominator as

This leads us to the formulas

and