10.2 Intersecting Chords, Secant/Tangent Lengths, Arc Length & Sector Area

10.2 Intersecting Chords, Secant/Tangent Lengths, Arc Length & Sector Area

Power of a Point: Geometric Foundations and Chord Products

In classical circle geometry, Jakob Steiner formulated the unified concept of the Power of a Point. For a given circle $\mathcal{C}(O, r)$ and any point $P$ in the plane at distance $d = OP$ from center $O$, the power of point $P$ with respect to the circle is defined as $\Pi(P) = d^2 - r^2$. When point $P$ lies inside the circle, $d < r$, producing a negative power; when $P$ lies outside, $d > r$, producing a positive power; and when $P$ lies on the boundary, $d = r$, producing zero power. Remarkably, whenever any line through $P$ intersects the circle at points $A$ and $B$, the product of the signed directed distances $PA \cdot PB$ is invariant and identically equals $d^2 - r^2$.

Intersecting Chords Theorem (Interior Point)

When two chords $\overline{AB}$ and $\overline{CD}$ intersect at an interior point $P$, the point partitions each chord into two positive segments. The theorem states that the product of the segments of one chord equals the product of the segments of the other: $AP \cdot PB = CP \cdot PD$

Deductive Proof via Triangle Similarity: Construct auxiliary chords $\overline{AD}$ and $\overline{CB}$. Consider triangles $\triangle APD$ and $\triangle CPB$:

1. $\angle APD \cong \angle CPB$ because vertical angles are congruent.
2. $\angle PDA \cong \angle PBC$ because both are inscribed angles intercepting the identical arc $\widehat{AC}$.

By Angle-Angle (AA) similarity, $\triangle APD \sim \triangle CPB$. Writing the ratio of corresponding side lengths yields: $\frac{AP}{CP} = \frac{PD}{PB} \implies AP \cdot PB = CP \cdot PD$ This fundamental relation allows any unknown segment length to be solved algebraically given the remaining three segments.

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External Metric Power: Secant-Secant and Tangent-Secant Theorems

When point $P$ lies outside the circle, lines drawn from $P$ can intersect the circle twice (secants) or once (tangents). The invariant power of a point translates into two primary metric formulas:

Secant-Secant Theorem

Let two secant lines be drawn from external point $P$. The first secant intersects the circle at $A$ (near) and $B$ (far), while the second secant intersects at $C$ (near) and $D$ (far). The theorem asserts: $PA \cdot PB = PC \cdot PD \quad \Longleftrightarrow \quad (\text{external}_1) \cdot (\text{whole}_1) = (\text{external}_2) \cdot (\text{whole}_2)$

Proof: Draw auxiliary chords $\overline{AD}$ and $\overline{BC}$. In triangles $\triangle PAD$ and $\triangle PCB$:

1. $\angle P \cong \angle P$ by the reflexive property of congruence.
2. Inscribed angle $\angle PDA$ intercepts arc $\widehat{AC}$, and inscribed angle $\angle PBC$ intercepts the same arc $\widehat{AC}$. Thus, $\angle PDA \cong \angle PBC$.

By AA similarity, $\triangle PAD \sim \triangle PCB$. Setting up the ratio of corresponding sides: $\frac{PA}{PC} = \frac{PD}{PB} \implies PA \cdot PB = PC \cdot PD$ Crucial Examinee Warning: A pervasive error on certification examinations is multiplying the external segment by the internal chord segment ($PA \cdot AB$). The theorem strictly requires multiplying the external segment by the entire secant length ($PA \cdot PB = PA \cdot (PA + AB)$).

Tangent-Secant Theorem

If a line from external point $P$ is tangent to the circle at point $T$, and a secant line from $P$ intersects the circle at near point $A$ and far point $B$, the relationship becomes: $PT^2 = PA \cdot PB \quad \Longleftrightarrow \quad (\text{tangent})^2 = (\text{external}) \cdot (\text{whole})$

Proof: Draw segments $\overline{TA}$ and $\overline{TB}$. In triangles $\triangle PTA$ and $\triangle PBT$, angle $\angle P$ is shared. By the Tangent-Chord Theorem, $m(\angle PTA) = \frac{1}{2}m(\widehat{TA})$. Inscribed angle $\angle PBT$ also intercepts arc $\widehat{TA}$, so $m(\angle PBT) = \frac{1}{2}m(\widehat{TA})$. Hence, $\angle PTA \cong \angle PBT$. By AA similarity, $\triangle PTA \sim \triangle PBT$: $\frac{PT}{PB} = \frac{PA}{PT} \implies PT^2 = PA \cdot PB$

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Tangent Properties, Chord Bisectors, and Center Orthogonality

Two-Tangent Theorem (Tangent Segments from an External Point)

When two tangent segments $\overline{PT_1}$ and $\overline{PT_2}$ are drawn to a circle from a single external point $P$, the tangent segments are congruent: $PT_1 = PT_2$ Proof: Construct radii $\overline{OT_1}$ and $\overline{OT_2}$ and segment $\overline{OP}$. Because tangents are perpendicular to radii at their points of contact, $\angle OT_1 P = \angle OT_2 P = 90^\circ$. In right triangles $\triangle OT_1 P$ and $\triangle OT_2 P$, hypotenuse $\overline{OP}$ is shared, and legs $OT_1 = OT_2 = r$. By the Hypotenuse-Leg (HL) Congruence Theorem, $\triangle OT_1 P \cong \triangle OT_2 P$. Therefore, $PT_1 = PT_2$ by CPCTC.

Chord Perpendicular Bisector Theorems

The relationship between a circle's center, chords, and perpendicular lines provides essential coordinate and synthetic tools:

1. Bisector Passes Through Center: The perpendicular bisector of any chord passes through the center of the circle. Consequently, the center of a circle circumscribing any polygon is the circumcenter—the concurrency point of the perpendicular bisectors of its sides.
2. Orthogonal Diameter Bisects Chord: A diameter or radius perpendicular to a chord bisects the chord and its intercepted arcs.
3. Pythagorean Metric for Chord Distance: If a chord of total length $c$ is at perpendicular distance $d$ from the center in a circle of radius $r$, the center, the chord midpoint, and an endpoint form a right triangle: $d^2 + \left(\frac{c}{2}\right)^2 = r^2$ This equation directly relates chord length, radius, and center-to-chord distance.

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Circular Measurement: Arc Length, Sector Area, and Circular Segments

Radian Measure and Arc Length

A radian is the plane angle subtended at the center of a circle by an arc whose length is equal to the radius of the circle: $\theta = \frac{s}{r}$. Because the circumference of a circle of radius $r$ is $C = 2\pi r$, a full revolution contains $\frac{2\pi r}{r} = 2\pi$ radians, establishing the conversion factor $180^\circ = \pi \text{ radians}$.

The linear length of an intercepted arc $s$ subtended by central angle $\theta$ is:

• In radians: $s = r\theta$
• In degrees: $s = \frac{\theta}{360^\circ}(2\pi r) = \frac{\pi r \theta}{180^\circ}$

Sector Area

A circular sector is the pie-shaped portion of a circle bounded by two radii and the intercepted arc. Its area is proportional to the fraction of the total circular area $\pi r^2$:

• In radians: $A_{\text{sector}} = \frac{1}{2}r^2\theta$
• In degrees: $A_{\text{sector}} = \frac{\theta}{360^\circ}(\pi r^2)$

Area of a Circular Segment

A circular segment is the region bounded by a chord and its intercepted arc. The area of a minor circular segment is obtained by subtracting the area of the isosceles triangle formed by the two radii and the chord from the total sector area: $A_{\text{segment}} = A_{\text{sector}} - A_{\triangle}$ Using the SAS triangle area formula $A_{\triangle} = \frac{1}{2}r^2 \sin\theta$:

• In radians: $A_{\text{segment}} = \frac{1}{2}r^2(\theta - \sin\theta)$
• In degrees: $A_{\text{segment}} = \frac{\theta}{360^\circ}(\pi r^2) - \frac{1}{2}r^2 \sin\theta$

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Power of a Point and Segment Metric Formulations Reference Table

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Worked Computational Exemplar: Exact Circular Segment Area Evaluation

Problem: A circle has radius $r = 12\text{ cm}$. A central angle $\theta = 120^\circ$ subtends chord $\overline{AB}$. Determine the exact length of chord $\overline{AB}$, the exact area of sector $OAB$, and the exact area of the circular segment bounded by chord $\overline{AB}$ and arc $\widehat{AB}$.

• Step 1: Calculate Chord Length $\overline{AB}$: Drop an altitude from center $O$ to chord $\overline{AB}$, bisecting the $120^\circ$ central angle into two $60^\circ$ angles and bisecting chord $\overline{AB}$ into two congruent segments of length $x$. Each half forms a $30^\circ-60^\circ-90^\circ$ right triangle with hypotenuse $r = 12$: $x = r \sin(60^\circ) = 12 \left(\frac{\sqrt{3}}{2}\right) = 6\sqrt{3}\text{ cm}$ $AB = 2x = 2(6\sqrt{3}) = 12\sqrt{3}\text{ cm}$
• Step 2: Calculate Sector Area: $A_{\text{sector}} = \frac{\theta}{360^\circ}\pi r^2 = \frac{120^\circ}{360^\circ} \pi (12^2) = \frac{1}{3} (144\pi) = 48\pi\text{ cm}^2$
• Step 3: Calculate Central Triangle Area: $A_{\triangle} = \frac{1}{2} r^2 \sin(120^\circ) = \frac{1}{2}(144)\left(\frac{\sqrt{3}}{2}\right) = 72 \left(\frac{\sqrt{3}}{2}\right) = 36\sqrt{3}\text{ cm}^2$ Alternatively, using base $AB = 12\sqrt{3}$ and altitude $d = r\cos(60^\circ) = 12(1/2) = 6$: $A_{\triangle} = \frac{1}{2}(12\sqrt{3})(6) = 36\sqrt{3}\text{ cm}^2$
• Step 4: Calculate Circular Segment Area: Subtract triangle area from sector area: $A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} = 48\pi - 36\sqrt{3}\text{ cm}^2$