The power of a point
Tonight's idea
Fix a circle and a point \(P\). Draw a line through \(P\) that cuts the circle at \(A\) and \(B\). Draw another that cuts it at \(C\) and \(D\).
Power of a point. \(PA \cdot PB = PC \cdot PD\). The product is the same for every line through \(P\).
\(P\) inside: two chords. Look at triangles \(PAC\) and \(PDB\). Their angles at \(P\) are equal, because they are opposite angles where two lines cross. Their angles at \(A\) and \(D\) are equal, because both stand on the arc \(BC\). That is "angles in the same segment" from Circle theorems, with proofs. Two equal angles make the triangles similar, so matching sides are in the same ratio:
\[\frac{PA}{PD} = \frac{PC}{PB} \quad\Longrightarrow\quad PA \cdot PB = PC \cdot PD.\]
\(P\) outside: two secants. A secant is a line that cuts the circle twice. Use triangles \(PAD\) and \(PCB\). They share the angle at \(P\). Their angles at \(D\) and \(B\) are equal, because both stand on the arc \(AC\). Similar again:
\[\frac{PA}{PC} = \frac{PD}{PB} \quad\Longrightarrow\quad PA \cdot PB = PC \cdot PD. \qquad \square\]
Swing a secant round until \(A\) and \(B\) merge. It becomes a tangent touching at \(T\), and the product becomes \(PT^2\).
Watch one: what the constant is
Chords \(AB\) and \(CD\) cross at \(P\), with \(PA = 3\), \(PB = 8\) and \(PC = 4\). Then \(3 \times 8 = 4 \times PD\), so \(PD = 6\).
Now suppose the radius is \(r = 7\). How far is \(P\) from the centre \(O\)? Call that distance \(d\) and use the one line we have not tried: the diameter through \(P\). It is cut into pieces \(r - d\) and \(r + d\), so
\[(r - d)(r + d) = r^2 - d^2 = 24 \quad\Longrightarrow\quad d^2 = 49 - 24 = 25, \quad d = 5.\]
So inside the circle the constant is \(r^2 - d^2\). Outside, the pieces are \(d - r\) and \(d + r\), and the constant is \(d^2 - r^2\).
Your turn
- Chords \(AB\) and \(CD\) cross at \(P\) inside a circle. \(PA = 4\), \(PB = 9\) and \(PC = 3\). Find \(PD\).
- From a point \(P\) outside a circle, one secant meets the circle at \(A\) and then \(B\), with \(PA = 4\) and \(AB = 5\). A second secant meets it at \(C\) and then \(D\), with \(PC = 3\). Find the length of the chord \(CD\).
- From \(P\) outside a circle, a secant meets the circle at distances 4 and 9 from \(P\). Find the length of the tangent \(PT\). If the radius is 8, how far is \(P\) from the centre?
- \(AB\) is a diameter. \(P\) lies on it with \(PA = 2\) and \(PB = 8\). The chord through \(P\) at right angles to \(AB\) is cut in half by \(P\). Find the length \(h\) of each half. Now replace 2 and 8 by any positive \(a\) and \(b\). Explain why \(h\) cannot be longer than the radius, and write down the inequality that this gives.
Stretch: two circles at once
Two circles cross at \(X\) and \(Y\). Take any point \(P\) on the line \(XY\) that is outside both circles. Show that the tangent from \(P\) to one circle has the same length as the tangent from \(P\) to the other.
Answers
- \(4 \times 9 = 3 \times PD\), so \(PD = 12\).
- \(PB = 4 + 5 = 9\), so the product is \(4 \times 9 = 36\). Then \(3 \times PD = 36\) gives \(PD = 12\), and \(CD = PD - PC = 9\).
- \(PT^2 = 4 \times 9 = 36\), so \(PT = 6\). Outside the circle the constant is \(d^2 - r^2\): \(d^2 - 64 = 36\), so \(d = 10\).
- The two halves of the chord multiply to \(PA \cdot PB\): \(h \times h = 2 \times 8\), so \(h = 4\). In general \(h^2 = ab\), so \(h = \sqrt{ab}\). The diameter is \(a + b\), so the radius is \(\tfrac{a+b}{2}\). No chord is longer than a diameter, so half a chord is at most a radius: \(\sqrt{ab} \le \tfrac{a + b}{2}\). They are equal only when \(P\) is the centre, that is when \(a = b\). (Here \(4 \le 5\).)
Stretch. The line \(XY\) is a secant of both circles, and it meets each of them at \(X\) and \(Y\). So the power of \(P\) is \(PX \cdot PY\) for the first circle and \(PX \cdot PY\) for the second. Each tangent length squared equals that same number, so the two tangents are equal.