It is a general pattern that if one is given 3 objects, each of which is a point or a circle, then there is exactly one circle that either passes through (when the object is a point) or is orthogonal to (when the object is a circle) each of the 3 objects.
The detailed decription below works through all the cases, but with some reflection it is easy to fit them all into a pattern. However, despite the similarity of approach, one should actually carry out all these constructions for various arrangements of the points and circles until the method is nailed down.
On the other hand, some of the special cases below should probably wait until the second pass through the ideas and not be a distraction the first time through.
Given THREE points and ZERO circles, construct a circle d through the points and orthogonal to the circles.
If the 3 points A, B, C are not collinear, then this is just the circumcircle of the triangle ABC. The center P of d is the point of concurrence of the three perpendicular bisectors of AB, BC, CA. P can be constructed by intersecting any 2 of the perpendicular bisectors. Notice the connection with the other cases in this list -- the perpendicular bisector of two points is the radical axis of the two points if they are considered "circles of radius zero".
The radius of the circle d is just PA (or PB or PC).
Special Case. If the 3 points are collinear, there is then a line through the 3 points but the perpendicular bisectors are parallel and there is no circle through them.
Given TWO points and ONE circle, construct a circle d through the points and orthogonal to the circles.
Given two points A and B and a circle c, then the center P of the circle d is the point of concurrence of the perpendicular bisector of AB, and the radical axes of A and c and of B and c. The radical axis of A and c can be constructed as the perpendicular bisector of AA', where A' is the inversion of A in c (unless A is on c, when the radical axis is the tangent to c at A). The radical axis of B and c is constructed the same way from the inversion B' of B in c.
Thus the center P is the circumcenter of the triangle ABA' or the triangle ABB' or even the triangle AA'B'. P can be constructed, among other ways as
- the intersection of the radical axis of A and c with the radical axis of B and c, or
- the intersection of the radical axis of A and c with the perpendicular bisector of AB, or
- the intersection of the perpendicular bisector of AB with the perpendicular bisector of A'B', etc.
The radius of d is PA or PB (or PA' or PB'). The radius is also the length of a tangent from P to c.
Special Case 1. If the points A and B and the center of c are collinear, then the intersection P does not exist, but the line AB is a line orthogonal to c, so it may be counted as a "circle of infinite radius".
Special Case 2. If B = A', than any circle through AB satisfies the conditions, so there are an infinite number of solutions.
Given ONE point and TWO circles, construct a circle d through the points and orthogonal to the circles.
Given a point A and two circles b and c, the center P of the circle d is the point of concurrence of these 3 lines: the radical axis of A and b, radical axis of A and c, and radical axis of b and c.
Thus P can be constructed as the intersection of two radical axes that are the axes of a point and a circle as above. The construction of the radical axis of two circles is consider in Assignment 1.
The radius of d is PA (or PA'). The radius is also the length of a tangent from P to b or from P to c.
Special Case 1. If the point A and the centers of b and c are collinear, then the intersection P does not exist, but the line through A and the centers is a line orthogonal to both b and c, so it may be counted as a "circle of infinite radius". If A and the two centers of b and c are all the same point, then there is an infinite number of these lines, but in other cases, there is only one line.
Special Case 2. If A is not on b or c, but the inversion of A in b and the inversion of A in c are the same point A', then any circle through AA' satisfies the conditions, so there are an infinite number of solutions. This can happen in exactly one way when b and c are disjoint. See the discussion of Appolonian circles and orthgonal pencils.
Special Case 3. If A is on the intersection of b and c, then if b and c are tangent at A, any circle through A orthogonal to b is a solution to the problem, so there are an infinite number of circles. If b and c are not tangent, then any circle through A orthogonal to b will not be orthogonal to c so there are no solutions.
Given ZERO points and THREE circles, construct a circle d through the points and orthogonal to the circles.
Given three circles a, b and c, the center P of the circle d is the point of concurrence of these 3 lines: the radical axis of a and b, the radical axis of b and c, and the radical axis of c and a. This point P is called the radical center of the 3 circles.
Thus P can be constructed as the intersection of two radical axes. The construction of the radical axis of two circles is consider in Assignment 1.
The radius of d the length of a tangent from from P to a or from P to b or from P to c. All these tangent lengths are the same since P is on all the radical axes.
Special Case 1. If the centers of a, b and c are collinear, then the intersection P does not exist, but the line through A and the centers is a line orthogonal to both b and c, so it may be counted as a "circle of infinite radius". If the three centers of a, b and c are all the same point, then there is an infinite number of these lines, but in other cases, there is only one line.
Special Case 2. If in addition to having collinear centers, the circles a, b, and c all belong to the same pencil, then there is an infinite set of circles (the orthogonal pencil) orthogonal to a, b, c. See the discussion of pencils and orthogonal pencils.