Quadrilateral In a Circle Consider a quadrilateral ABCD inscribed in a circle as shown below. The point O is the center of the circle, and we let a, b, g denote the angles AOD, DOC, and AOB respectively.  The two diagonals of the quadrilateral meet at the point I.  The distance from O to I is denoted by s.  If we define  m = (a - b)/2  and  n = (g - a)/2, we have The diagonals are perpindicular if and only if g + b = p. An example of this circumstance is illustrated below. With perpendicular diagonals, we have g = p - b, so the above expressions reduce to Of course, the area of the quadrilateral is sin(a) + sin(b). Also in the case of perpindicular diagonals, suppose we take these two diagonals as the diameters of two new circles. Each of these circles intersects the other diagonal in two points. These four points of intersection form a perfect square, with area 2sin(a)sin(b), as illustrated below. This is a fairly immediate corrollary of Pythagoras's theorem.  If AC is the hypotenuse of a right triangle ABC, then the point B is on the circle whose diameter is AC, as shown below. (This can be proven simply by completing the rectangle.) The conventional statement of "right angle relations" is the one given by Pythagoras and Euclid, which is However, another necessary and sufficient condition for right angles is which is the hyperbolic expression of Pythagoras's theorem. This signifies that the height of a right triangle is the geometric mean of the two segments formed by dropping a perpendicular from the apex to the base. Hence the right triangles drawn on the two diagonals of the original quadrilateral have the same heights, resulting in a perfect square. Of course (2) is really just a special case of a more general relation.  (See Propositions 35 and 36 of Book III of Euclid's Elements.)  Suppose we have a circle of radius r and an arbitrary point on the plane (inside or outside the circle) called P.  Draw any line through the point P, and let S,T denote the points of intersection with the circle, as shown below. Then we have where R is the distance from P to the center of the circle. Obviously if TS is a diameter of the circle then r2 - R2 is just the squared height of the perpindicular at P, and equation (3) reduces to (2).  However, the nice thing about the more general form (3) is that it works for any line through the circle, not just diameters.  Also, note that for any line through the point P the product of the two segments is the same (because the distance from P to the center of the circle is unchanged). The same theorem applies even when the point P is outside the circle, as shown below. where again R is the distance from P to the center of the circle. In fact, this even works when the line through P does not intersect the circle.  In that case the points of intersection have complex coordinates in the plane, and the distances |PT| and |PS| are complex, but they are conjugates, and their product is still the real number  R2 - r2. To prove this algebraically, place a circle of radius r on the x axis of Cartesian coordinates, with its center a distance R from the origin A, and draw an arbitrary line through the origin intersecting the circle at points B and C as shown below. The equation of the circle is and the equation of the line AC is Substituting into the previous expression and collecting terms on x gives the quadratic Making use of the fact that the quadratic in x is which shows that the x coordinates of the two points of intersection B and C satisfy Since x1 = |AB| cos(a) and x2 = |AC| cos(a), it follows from the above relation that (Euclid’s proof of this proposition based on the Pythagorean theorem – along with a simpler proof based on calculus - are presented in Generalizing Pythagoras and Carnot.) Incidentally, the difference between the roots of the quadratic is so we have More generally, given any quadratic curve or surface centered at point Pc = (xc,yc,zc) consisting of the locus of points that satyisfy the equation where a, b, c, r are constants, we can define the "distance function" d(P1,P2) for any two points P1 and P2 by Now, for any given line through the origin there are constants ky and kz such that so we can find the points of intersection of the line with the conic locus by substituting into the locus equation and collecting terms to give the quadratic in x The product of the two roots x1 and x2 equals the last term. We also have the “distances” from the origin P0 to the two points of intersection P1 and P2 Therefore, we have Thus the general form of Pythagoras's theorem can be expressed as where the sign of d(u,v) is based on a consistent definition of polarity on the line from u to v.  Thus the signs of d(u,v) and d(v,u) are opposite.  Consequently, if the locus is a circle and the point P0 is outside the circle, then d(P1,P0) and d(P0,P2) have opposite signs, so the right hand term can be brought over to the left as a positive magnitude. The figure below illustrates the application of equation (4) to a hyperbolic locus. Moreover, if P1 = P2 then the line from P0 is tangent to the locus and equation (4) reduces to the familiar “Pythagorean” form The same proposition can be generalized to any homogeneous algebraic function of n variables. For example, consider the function Using this function we can define the “distance” between two points P1 and P2 by the equation Now consider the locus of points that are a “distance” r from a given central point Pc, i.e., the points satisfying the equation and also consider an arbitrary line through the origin P0, with the equations Substituting these expressions for y and z into the previous equation gives a cubic in x. The coefficient of x3 in this cubic is easily seen to be and the constant coefficient is Therefore, letting x1, x2, x3 denote the roots of the cubic in x, we know their product equals (-1)n times the constant coefficient divided by the coefficient of x3, so we have The cube of the left hand side can be expressed as so we have For the fully general case of an nth degree homogeneous polynomial function f of m variables, we can let P1, P2, …, Pn denote the n points of intersection of an arbitrary line through any given point P0 with the locus d(P,Pc) = rn, then we have the identity Return to MathPages Main Menu