Wilson Stothers' Inversive Geometry and CabriJava Pages

Orthogonal Curves

In angles between arcs, we looked at the signed angle between two arcs meeting at a point.
Here, we will consider the magnitude of the angle(s) between two curves which meet at a point.
As before, we define the angles in terms of the tangents at the point of intersection.
Suppose that, in the sketch, <AOC has magnitude a.
Remember that the magnitude of an angle lies in (0,p].
Then <BOD also has magnitude a, while <AOD and <BOC each have magnitude p - a.
Special case
Suppose that the angle a is a right angle. Then all four angles round O are right angles.
Definition Two intersecting curves are orthogonal if they meet at right angles.

Orthogonal lines and circles
Obviously, two lines are orthogonal if and only if they are perpendicular.
Recall that, for a circle C, centre O, the tangent to C at a point P on C
is the line through P perpendicular to OP.
Conversely, the line through P perpendicular to the tangent at P is a diameter of C.
In the CabriJava pane, you can drag P and watch the effect on
the angle between the radius OP and the tangent (the green line).

Now consider the case of two orthogonal circles C and D,
with centres P and Q, meeting at A and B.
Let L be the tangent to C at A, and M the tangent to D at A.
Since the circles are orthogonal, L and M are perpendicular.
As M is perpendicular to the tangent M to C,
it passes through P, the centre of C.
Similarly, L passes through Q, the centre of D.
This is illustrated in the CabriJava pane on the right.
Observe that the tangent to C at B must also pass through Q.
Thus the choice of A and B on C determines D
provided the tangents at A and B meet.
If the tangents do not meet, then they must be parallel, so AB is a diameter of C.
Thus we have
The Uniqueness Theorem for Circles
If points A and B lie on a circle C, then there is a unique i-line
through A and B orthogonal to C.
As you may guess, we can replace C by an extended line (so one of A, B may be Ñ).
The result still holds, but again we need to take various cases.
A better approach to this and similar problems is to apply inversion
to reduce the number of cases.

The Uniqueness Theorem
If A and B are distinct points on an i-line L, then there is a unique i-line M
through A and B orthogonal to L.

Since the proof is a good example of
the use of inversion, we give it in full.

Proof
Since A and B are distinct, one of them (A say) is not Ñ.
Let C be a circle with centre A, so L passes through the centre of C.
Let M be any i-line through A and B.
Let B', L', M' be the inverses of B, L, M with respect to C.
By the Invariance Theorem, L and M are orthogonal if and only if
L' and M' are.
By the Inversion Theorem, L' and M' are extended lines.
As B is on L and M, the inverses pass through B'.
But there is exactly one i-line, N say, through B' perpendicular to L'.
Hence there is a unique M (the inverse of N with respect to C).
Note the structure of the proof.

We transform the picture to get a single, easy case.
We verify that the result in the special case
implies the result in the original picture.
To proceed further, it is convenient to introduce complex coordinates.
We then get an algebraic version of the Invariance Theorem.
This gives much more information about inversion.

Main Inversive Page

In angles between arcs, we looked at the signed angle between two arcs meeting at a point. Here, we will consider the magnitude of the angle(s) between two curves which meet at a point. As before, we define the angles in terms of the tangents at the point of intersection. Suppose that, in the sketch, <AOC has magnitude a. Remember that the magnitude of an angle lies in (0,p]. Then <BOD also has magnitude a, while <AOD and <BOC each have magnitude p - a. Special case Suppose that the angle a is a right angle. Then all four angles round O are right angles. Definition Two intersecting curves are orthogonal if they meet at right angles.
Orthogonal lines and circles Obviously, two lines are orthogonal if and only if they are perpendicular. Recall that, for a circle C, centre O, the tangent to C at a point P on C is the line through P perpendicular to OP. Conversely, the line through P perpendicular to the tangent at P is a diameter of C. In the CabriJava pane, you can drag P and watch the effect on the angle between the radius OP and the tangent (the green line).
Now consider the case of two orthogonal circles C and D, with centres P and Q, meeting at A and B. Let L be the tangent to C at A, and M the tangent to D at A. Since the circles are orthogonal, L and M are perpendicular. As M is perpendicular to the tangent M to C, it passes through P, the centre of C. Similarly, L passes through Q, the centre of D. This is illustrated in the CabriJava pane on the right. Observe that the tangent to C at B must also pass through Q. Thus the choice of A and B on C determines D provided the tangents at A and B meet. If the tangents do not meet, then they must be parallel, so AB is a diameter of C. Thus we have The Uniqueness Theorem for Circles If points A and B lie on a circle C, then there is a unique i-line through A and B orthogonal to C. As you may guess, we can replace C by an extended line (so one of A, B may be Ñ). The result still holds, but again we need to take various cases. A better approach to this and similar problems is to apply inversion to reduce the number of cases.
The Uniqueness Theorem If A and B are distinct points on an i-line L, then there is a unique i-line M through A and B orthogonal to L.	Since the proof is a good example of the use of inversion, we give it in full.
Proof Since A and B are distinct, one of them (A say) is not Ñ. Let C be a circle with centre A, so L passes through the centre of C. Let M be any i-line through A and B. Let B', L', M' be the inverses of B, L, M with respect to C. By the Invariance Theorem, L and M are orthogonal if and only if L' and M' are. By the Inversion Theorem, L' and M' are extended lines. As B is on L and M, the inverses pass through B'. But there is exactly one i-line, N say, through B' perpendicular to L'. Hence there is a unique M (the inverse of N with respect to C). Note the structure of the proof. We transform the picture to get a single, easy case. We verify that the result in the special case implies the result in the original picture. To proceed further, it is convenient to introduce complex coordinates. We then get an algebraic version of the Invariance Theorem. This gives much more information about inversion.