Two Secants
A secant of a circle is a line connecting two points on the circle. When two nonparallel secants are drawn, a number of useful properties are satisfied, even if the two intersect outside the circle.
These properties are especially useful in the context of cyclic quadrilaterals, as they often allow various angles and/or lengths to be filled in. In fact, these results are so useful that it is not unusual to add lines to a diagram for the purpose of creating two-secant configurations.
Intersection inside the circle
When the two secants intersect inside the circle, they can be viewed as diagonals of a cyclic quadrilateral, and as such they satisfy the same properties. Let \(AB\) and \(CD\) be the two secants, intersecting inside the circle at point \(P\). The first major property is that pairs of angles subtending the same arc are equal, as per the inscribed angle theorem, which gives four important angle equalities:
\(\angle ABD = \angle ACD\), since they both subtend arc \(\overset{\frown}{AD}\). Similarly, \(\angle ABC = \angle ADC\), \(\angle CAB = \angle CDB\), and \(\angle BAD = \angle BCD\), by looking in turn at the arcs \(\overset{\frown}{AD}, \overset{\frown}{AC}, \overset{\frown}{CB},\) and \(\overset{\frown}{BD}\).
A similar property is that opposite angles add to \(180^{\circ}\), again due to the inscribed angle properties.
\[\angle ACB + \angle ADB = \angle BAD + \angle BCD = 180^{\circ}\]
The final important property is the first case of the Power of a Point theorem, which states
\[PA \cdot PB = PC \cdot PD\]
Intersection outside the circle
\(\angle ABC = \angle ADC\), since they both subtend arc \(\overset{\frown}{AC}\). Similarly, \(\angle CAD = \angle CBD\), \(\angle DAB = \angle DCB\), and \(\angle BCA = \angle BDA\), by looking in turn at the arcs \(\overset{\frown}{AC}, \overset{\frown}{CD}, \overset{\frown}{DB},\) and \(\overset{\frown}{BA}\).
As shown in the diagram, \(E\) and \(F\) form a circle with two of the vertices of the quadrilateral (say \(B\) and \(D\)).
Find the angle between the lines \(EF\) and \(AC\).
This problem is part of the Advent Calendar 2015.
Again, opposite angles add to \(180^{\circ}\):
\[\angle ABC + \angle ADC = \angle CAB + \angle CDB = 180^{\circ}\]
The final important property is the second case of the Power of a Point theorem, which states
\[ PA \cdot PB = PC \cdot PD\]
Note that this is the same statement as the previous section. In fact, the power of a point with respect to a circle depends only on its distance to the origin; it doesn't matter whether the point is inside or outside the circle. Specifically, the power of point \(P\) is \(OP^2-r^2\), where \(r\) is the radius of circle \(O\), and this is independent of the secant chosen.
Consider the circle \(\Gamma\) whose equation is \(x^2 + y^2- 28x + 40y + 20 = 0\). Let \(S\) be the set of all points \(P\) outside \(\Gamma\) such that if there is a line through \(P\) which intersects \(\Gamma\) at two points \(A\) and \(B\), then \(PA\cdot PB = 100\). Find the minimum possible distance between a point on \(\Gamma\) and a point on \(S\).