GeometryDifficulty 35-85448 tagged problems

Power of a Point

Power of a Point is the habit of relating secants and tangents through equal products from a fixed point and a circle. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\text{For a point } P \text{ and a circle: } PA\cdot PB = PC\cdot PD = |PT|^2 = |\,d^2 - r^2|$$

From a fixed point, the product of the two distances along any line through a circle is the same, whether the lines are secants or a tangent.

Why It Works

1
Draw two chords through $P$ meeting the circle at $A,B$ and $C,D$. Triangles $\triangle PAC$ and $\triangle PDB$ share angle $\angle P$ and have equal inscribed angles ($\angle PAC = \angle PDB$), so they are similar.
2
Similarity gives $\dfrac{PA}{PD} = \dfrac{PC}{PB}$, i.e. $PA\cdot PB = PC\cdot PD$ — the product is independent of which line you pick.
3
If $P$ is outside and a line is tangent at $T$, the limiting case $C \to D \to T$ gives $PA\cdot PB = PT^2$.
4
In terms of the distance $d$ from $P$ to the center and radius $r$, this common value equals $d^2 - r^2$ (negative inside the circle, where chords cross).

Worked Examples

Example 1

From an external point $P$, a tangent touches a circle at $T$ with $PT = 6$. A secant from $P$ meets the circle at $A$ (near) and $B$ (far) with $PA = 4$. Find $AB$.

1
Power of a point gives $PT^2 = PA\cdot PB$, so $6^2 = 4\cdot PB$.
2
Then $PB = \dfrac{36}{4} = 9$.
3
Since $A$ is between $P$ and $B$, $AB = PB - PA = 9 - 4 = 5$.

Answer:

$AB = 5$

Warm-up

Two chords of a circle, $AB$ and $CD$, intersect at an interior point $P$. If $PA = 3$, $PB = 8$, and $PC = 4$, find $PD$.

1
For two chords crossing inside a circle, power of a point gives $PA\cdot PB = PC\cdot PD$.
2
Substitute: $3\cdot 8 = 4\cdot PD$, so $24 = 4\cdot PD$.
3
Therefore $PD = 6$.

Answer:

$PD = 6$

Contest level

A circle has radius $10$ and center $O$. A point $P$ is at distance $OP = 6$ from the center. A chord through $P$ is bisected by $P$. Find the length of that chord.

1
The power of the interior point $P$ is $OP^2 - r^2 = 6^2 - 10^2 = 36 - 100 = -64$, so for any chord through $P$ the (signed) product $PA\cdot PB = 64$ in magnitude.
2
If $P$ bisects the chord then $PA = PB = h$, so $h^2 = 64$ and $h = 8$.
3
The full chord is $2h = 16$. (Check: the shortest chord through $P$ is the one perpendicular to $OP$, of half-length $\sqrt{r^2 - OP^2} = \sqrt{100-36} = 8$ — consistent.)

Answer:

The chord has length $16$.

Olympiad / Challenge

In $\triangle ABC$, the altitudes meet at the orthocenter $H$. The altitude from $A$ meets $BC$ at $D$ and the altitude from $B$ meets $AC$ at $E$. Using power of a point at $H$, show $HA\cdot HD = HB\cdot HE$.

1
Points $D$ and $E$ are feet of altitudes, so $\angle BDH = \angle AEH = 90^\circ$; thus $A, B, D, E$ lie on the circle with diameter $AB$ (each of $D, E$ sees $AB$ at a right angle).
2
Lines $AD$ and $BE$ are two chords of THAT circle (extended) meeting at $H$.
3
Power of the point $H$ with respect to this circle gives $HA\cdot HD = HB\cdot HE$ (the products along the two chords through $H$ are equal).

Answer:

$HA\cdot HD = HB\cdot HE$ (equal power of $H$ on the circle through $A, B, D, E$).

Going Deeper

Generalization: the power of $P$ is the single signed number $\text{pow}(P) = OP^2 - r^2$, NEGATIVE inside the circle, zero on it, positive outside. Every secant/tangent product through $P$ equals this number (up to sign), which is exactly why crossing-chords, secant-secant, and secant-tangent are one theorem, not three.
Where it appears: power of a point converts the messy 'segment products' on AIME geometry into a single equation, proves the radical-axis and radical-center theorems, and is the standard way to compute a length when a tangent or two secants emanate from a common point.
Pitfall: keep the orientation straight. For an external point both distances are measured from $P$ outward, so $PA\cdot PB$ uses the near AND far intersection (not $AB$). For tangents it is $PT^2$, not $2\cdot PT$. Mixing up $PB$ with the chord length $AB$ is the most common arithmetic slip.

Spot the Signal

Look for problems where the key step is relating secants and tangents through equal products from a fixed point and a circle.
You can describe the hard part as relating secants and tangents through equal products from a fixed point and a circle, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the diagram relation that calls for power of a point, then rewrite the givens around it.
Name the relation that makes Power of a Point legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Power of a Point just because the surface looks familiar; verify the required condition first.
Applying Power of a Point because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let power of a point reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is relating secants and tangents through equal products from a fixed point and a circle.

Back to techniques

GeometryDifficulty 35-85448 tagged problems

Power of a Point

Practice this technique Match my level

The Key Result

$$\text{For a point } P \text{ and a circle: } PA\cdot PB = PC\cdot PD = |PT|^2 = |\,d^2 - r^2|$$

From a fixed point, the product of the two distances along any line through a circle is the same, whether the lines are secants or a tangent.

Why It Works

1
Draw two chords through $P$ meeting the circle at $A,B$ and $C,D$. Triangles $\triangle PAC$ and $\triangle PDB$ share angle $\angle P$ and have equal inscribed angles ($\angle PAC = \angle PDB$), so they are similar.
2
Similarity gives $\dfrac{PA}{PD} = \dfrac{PC}{PB}$, i.e. $PA\cdot PB = PC\cdot PD$ — the product is independent of which line you pick.
3
If $P$ is outside and a line is tangent at $T$, the limiting case $C \to D \to T$ gives $PA\cdot PB = PT^2$.
4
In terms of the distance $d$ from $P$ to the center and radius $r$, this common value equals $d^2 - r^2$ (negative inside the circle, where chords cross).

Worked Examples

Example 1

From an external point $P$, a tangent touches a circle at $T$ with $PT = 6$. A secant from $P$ meets the circle at $A$ (near) and $B$ (far) with $PA = 4$. Find $AB$.

1
Power of a point gives $PT^2 = PA\cdot PB$, so $6^2 = 4\cdot PB$.
2
Then $PB = \dfrac{36}{4} = 9$.
3
Since $A$ is between $P$ and $B$, $AB = PB - PA = 9 - 4 = 5$.

Answer:

$AB = 5$

Warm-up

Two chords of a circle, $AB$ and $CD$, intersect at an interior point $P$. If $PA = 3$, $PB = 8$, and $PC = 4$, find $PD$.

1
For two chords crossing inside a circle, power of a point gives $PA\cdot PB = PC\cdot PD$.
2
Substitute: $3\cdot 8 = 4\cdot PD$, so $24 = 4\cdot PD$.
3
Therefore $PD = 6$.

Answer:

$PD = 6$

Contest level

A circle has radius $10$ and center $O$. A point $P$ is at distance $OP = 6$ from the center. A chord through $P$ is bisected by $P$. Find the length of that chord.

1
The power of the interior point $P$ is $OP^2 - r^2 = 6^2 - 10^2 = 36 - 100 = -64$, so for any chord through $P$ the (signed) product $PA\cdot PB = 64$ in magnitude.
2
If $P$ bisects the chord then $PA = PB = h$, so $h^2 = 64$ and $h = 8$.
3
The full chord is $2h = 16$. (Check: the shortest chord through $P$ is the one perpendicular to $OP$, of half-length $\sqrt{r^2 - OP^2} = \sqrt{100-36} = 8$ — consistent.)

Answer:

The chord has length $16$.

Olympiad / Challenge

1
Points $D$ and $E$ are feet of altitudes, so $\angle BDH = \angle AEH = 90^\circ$; thus $A, B, D, E$ lie on the circle with diameter $AB$ (each of $D, E$ sees $AB$ at a right angle).
2
Lines $AD$ and $BE$ are two chords of THAT circle (extended) meeting at $H$.
3
Power of the point $H$ with respect to this circle gives $HA\cdot HD = HB\cdot HE$ (the products along the two chords through $H$ are equal).

Answer:

$HA\cdot HD = HB\cdot HE$ (equal power of $H$ on the circle through $A, B, D, E$).

Going Deeper

Generalization: the power of $P$ is the single signed number $\text{pow}(P) = OP^2 - r^2$, NEGATIVE inside the circle, zero on it, positive outside. Every secant/tangent product through $P$ equals this number (up to sign), which is exactly why crossing-chords, secant-secant, and secant-tangent are one theorem, not three.
Where it appears: power of a point converts the messy 'segment products' on AIME geometry into a single equation, proves the radical-axis and radical-center theorems, and is the standard way to compute a length when a tangent or two secants emanate from a common point.
Pitfall: keep the orientation straight. For an external point both distances are measured from $P$ outward, so $PA\cdot PB$ uses the near AND far intersection (not $AB$). For tangents it is $PT^2$, not $2\cdot PT$. Mixing up $PB$ with the chord length $AB$ is the most common arithmetic slip.

Spot the Signal

Look for problems where the key step is relating secants and tangents through equal products from a fixed point and a circle.
You can describe the hard part as relating secants and tangents through equal products from a fixed point and a circle, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the diagram relation that calls for power of a point, then rewrite the givens around it.
Name the relation that makes Power of a Point legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Power of a Point just because the surface looks familiar; verify the required condition first.
Applying Power of a Point because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let power of a point reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is relating secants and tangents through equal products from a fixed point and a circle.