Theorem 3.15: Given a circle with center   (x0, y0)   and radius   r   and a point   (x1, y1),

if the point is outside the circle, there are exactly two lines through the point tangent to the circle,
if the point is on the circle, there is exactly one line through the point tangent to the circle and it is perpendicular to the line joining the point to the center of the circle, and
if the point is inside the circle, there are no lines through the point tangent to the circle.

Proof: In the picture, it looks as if we are assuming that the point is outside the circle, but the following proof will work in all three cases.

If we use the simpler formula for the coordinates of the points where a line intersects a circle from Theorem 3.3,

and

Being tangent means that the line intersects the circle at exactly one point, and in order for that to happen the expression under the radicals,

(m2 + 1)r2 - (y0 - mx0 - b)2 = 0

This is a quadratic in   m,   and we can use the quadratic formula to solve for   m.   Before we simplify we will need to note that   b,   the y-intercept of the line through   (x1, y1)   with slope   m,   is given by

b = y1 - mx1

by Theorem 1.3. Substitute

(m2 + 1)r2 - (y0 - mx0 - (y1 - mx1))2 = 0

remove parentheses and combine like terms in the second term.

(m2 + 1)r2 - ((y0 - y1) - m(x0- x1))2 = 0

m2r2 + r2 - (y0 - y1)2 + 2m(x0- x1)(y0 - y1) - m2(x0- x1)2 = 0

Combine like terms

(r2 - (x0 - x1)2)m2 + 2(x0- x1)(y0 - y1)m + r2 - (y0 - y1)2

and we can use the quadratic formula. Let's use the form

if

ax2 + bx + c = 0

then

This gives us

This will give us two values for   m,   so long as the quantity under the radical is positive, one value for   m   if the quantity under the radical is   0,   and no real values for   m   if the quantity under the radical is negative. Let us examine the quantity under the radical

((x0- x1)(y0 - y1))2 - (r2 - (x0 - x1)2)( r2 - (y0 - y1)2)

= ((x0- x1)(y0 - y1))2 - r4 + ((x0 - x1)2 + (y0 - y1)2)r2 - ((x0- x1)(y0 - y1))2

If we recall that

d2 = (x0 - x1)2 + (y0 - y1)2

this simplifies to

d2r2- r4

= r2(d2 - r2)

If the point is outside the circle, this will be positive, and there will be two values for   m,   and, hence, two lines through the point tangent to the circle. If the point is on the circle, this will be   0,   and there will be only one value for   m,   and, hence only one line through the point tangent to the circle. If the point is inside the circle, this will be negative, and there will be no real value for   m,   and hence no lines through the point tangent to the circle.

If the point is on the circle, and   d = r,   then we have

r2 - (x0 - x1)2

= (x0 - x1)2 + (y0 - y1)2 - (x0 - x1)2

= (y0 - y1)2

and, since the radical term in the formula for the slope is   0,

which is the negative reciprocal of the slope of the line between the point through which the line passes and the center of the circle. Since that point is the point of tangency, it follows that the tangent is perpendicular to the line from the center of the circle to the point of tangency, and we have another analytic proof of Theorem 3.7.

There is only one other issue with which we must deal. What if the denominator in the formula for the slope is   0?   The denominator is

r2 - (x0 - x1)2

If this were   0,   it would say that

depending on which side of   x0 - x1   lies.The formula for the slopes,

becomes

Of the two answers, one is zero over zero, and the other is something nonzero over zero. The one which is something nonzero over zero gives us an infinite slope suggesting a vertical line. The 0/0 is indeterminate, and we can't get any information out of it.

But we can take the suggestion to consider the vertical line through   (x1, y1).   Its equation is

x = x1

If we substitute this into the equation of the circle, we get,

(x1 - x0)2 + (y - y0)2 = r2

But since we are operating under the assumption that   (x0 - x1)2 = r2,   this becomes

r2 + (y - y0)2 = r2

or

y = y0

and   (x1, y0)   is the only point where the vertical line through   (x1, y1)   meets the circle. That says that the vertical line through   (x1, y1)   is tangent to the circle at   (x1, y0).   In this case, the line from the point of tangency to the center of the circle would be horizontal, and again the tangent would be perpendicular to the line from the circle to the point of tangency. Since we are dealing with a vertical line, we should not be surprised that slope arguments will not work.

So to complete the proof, we need only show that in this case, where   r = |x0 - x1|,   that in addition to the vertical line through   (x1, y1),   there is a second line through   (x1, y1),   which is tangent to the circle. Let   O = (x0, y0),   A = (x1, y1),   and the point of tangency between the vertical line through   (x1, y1)   and the circle,   B = (x1, y0).   Let   |OB| = r,   the radius of the circle,   |AB| = a,   and let   |OA| = d.

Note,   |x0 - x1| = r,   |y0 - y1| = a,   and r2 + a2 = d2.   Since   OAB   forms a triangle, by the triangle inequality, Theorem 3.5

d < r + a,

and, also by the triangle inequality, Theorem 3.5

a < d + r

so

a - r < d

Similarly

r < a + d

so

r - a < d.

Coclude

|r - a| < d

Thus the conditions of Theorem 3.13 are satisfied and the given circle intersects the circle centered at   (x1, y1)   with radius   a   at two point. We have seen that one of them is   B.   Let us call the other one   C.

We know that   OAB   is a right triangle with right angle at   B,   so   r,   a,   and   d   satisfy the Pythagorean Theorem 3.4.

r2 + a2 = d2

Since these are the lengths of the sides in triangle   OAC,   we conclude that   OAC   is also a right triangle,with right angle at   C,   also by the Pythagorean Theorem, Theorem 3.4, so   C   is foot of the center of the first circle in the line determined by   AC,   and is thus the closest point on that line , by Theorem 3.6, to the center of the circle. All other point on the line will then be farther away than   C   and outside the circle. Conclude that   AC   is tangent to the circle at   C.

next theorem (3.16)