Power of a Point is one of the most useful theorems in competition geometry. It relates the lengths of segments created when two lines intersect a circle β whether the intersection is inside the circle (two chords) or outside (two secants, or a secant and a tangent). One formula, three cases.
For a point P and a circle with center O and radius r:
\[ \text{Power of P} = OP^2 - r^2 \]
The key insight: for any line through P intersecting the circle at X and Y, the product $ PX \cdot PY $ is constant β it equals the power of P (up to sign).
If two chords AB and CD intersect at point P inside the circle:
\[ PA \cdot PB = PC \cdot PD \]
The products of the segments of each chord are equal.
Why: Triangles PAC and PDB are similar by AA (vertical angles + inscribed angles subtending the same arc).
If from point P outside the circle, two secants are drawn β one intersecting the circle at A and B, the other at C and D (with PA < PB, PC < PD):
\[ PA \cdot PB = PC \cdot PD \]
\(Whole secant \\times external segment is constant for both secants.\)
Important: PA is the entire length from P to the far intersection B, times the near segment PA. Wait β \(more precisely: external part \\times whole = external part \\times whole.\)
Standard notation: if the secant hits the circle first at A, then continues to B (farther from P):
$ PA_{outside} \times PB_{whole} = PC_{outside} \times PD_{whole} $
If from point P outside the circle, a tangent touches at T and a secant passes through A (near) and B (far):
\[ PT^2 = PA \cdot PB \]
The square of the tangent length equals the product of the entire secant and its external segment.
This is the limit case of the secant-secant theorem where the two intersection points coincide.
All three cases say the same thing: for a point P and a line through P cutting the circle at X and Y, the product $PX \cdot PY$ is constant (depends only on P and the circle, not the line).
The set of points with equal power with respect to two circles is a line called the radical axis.
Power of a Point β Intersecting Chords:
\[ PA \cdot PB = PC \cdot PD \]
Step 1: Plug in the known values.
$ 4 \times 9 = 6 \times PD $
Step 2: Solve for PD.
$ 36 = 6 \cdot PD $
$ PD = 6 $
Check: \(4 \\times 9 = 36, 6 \\times 6 = 36. Equal products.\) β
Answer: 6
Power of a Point β Tangent-Secant:
\[ PT^2 = PA \cdot PB \]
Step 1: PT = 12, PA = 8 (external segment from P to near intersection A).
PB = whole secant (from P through A to far intersection B).
Step 2: Plug in values.
$ 12^2 = 8 \times PB $
$ 144 = 8 \cdot PB $
$ PB = 18 $
Check: \(PA \\times PB = 8 \\times 18 = 144 = 12^{2} = PT^{2}.\) β
Note: The chord AB inside the circle would be PB - PA = 18 - 8 = 10.
Answer: 18
Power of a Point β Secant-Secant:
\[ PA \cdot PB = PC \cdot PD \]
\((external \\times whole = external \\times whole)\)
Step 1: Plug in known values.
PA = 5, PB = 12, PC = 4. Find PD.
$ 5 \times 12 = 4 \times PD $
Step 2: Solve.
$ 60 = 4 \cdot PD $
$ PD = 15 $
Check: \(5 \\times 12 = 60, 4 \\times 15 = 60. Equal.\) β
Answer: 15