Circle Relationships: Angles, Radii, and Chords - Geometry

In the diagram, $AB$ is a diameter of circle $O$ , and $CD$ is a chord perpendicular to $AB$ at point $E$ . If $AE = 16$ , $EB = 4$ , what is the length of $CD$ ?
Answer: 16 Since $AB$ is a diameter, $AB = AE + EB = 16 + 4 = 20$ , so the radius is 10. The center $O$ is at the midpoint of diameter $AB$ , so $AO = OB = 10$ . Since $AE = 16$ and $AO = 10$ , point $E$ is $6$ units from the center ( $OE = |AE - AO| = |16 - 10| = 6$ ). When a chord is perpendicular to a diameter, the diameter bisects the chord. Using the relationship for intersecting chords: $AE \cdot EB = CE \cdot ED$ . Since $CD$ is bisected at $E$ , let $CE = ED = x$ , so $x^2 = 16 \cdot 4 = 64$ , giving $x = 8$ . Therefore, $CD = CE + ED = 8 + 8 = 16$ . Choice A (8) is half the chord length. Choice B (12) might result from using $AE - EB = 12$ . Choice D (20) is the diameter length.
Refer to the figure below. In circle $O$ , $\overline{AB}$ is a diameter, and $C$ is a point on the circle such that $\overline{AC} = 8$ and $\overline{BC} = 15$ . What is the radius of circle $O$ ?
Answer: $8.5$ Since $\overline{AB}$ is a diameter, the inscribed angle $\angle ACB = 90°$ . Then $AB = \sqrt{8^2 + 15^2} = \sqrt{289} = 17$ , so the radius is $17/2 = 8.5$ . Distractor A is half of 15 (mistaking the longer leg for the diameter). C averages the legs. D is the diameter, not the radius.
In circle $P$ , two secants are drawn from external point $R$ . Secant $RST$ has $RS = 9$ and $ST = 7$ . Secant $RUV$ has $RU = 6$ and $UV = x$ . What is the value of $x$ ?
Answer: 18 By the secant-secant theorem, when two secants are drawn from the same external point: $RS \cdot RT = RU \cdot RV$ , where $RT = RS + ST$ and $RV = RU + UV$ . So: $RT = 9 + 7 = 16$ and $RV = 6 + x$ . The equation becomes: $9 \cdot 16 = 6 \cdot (6 + x)$ , so $144 = 6(6 + x) = 36 + 6x$ , giving $6x = 108$ and $x = 18$ . Choice A (8) might result from using $RS - RU + ST = x$ . Choice B (12) could come from incorrectly using $\frac{RS \cdot ST}{RU} = x$ . Choice D (21) might result from using $RS + ST + RU - 1 = x$ .
In circle $O$ , tangent $PT$ touches the circle at point $T$ , and secant $PAB$ passes through the circle with $A$ and $B$ on the circle. If $PT = 12$ and $PA = 8$ , what is the length of $PB$ ?
Answer: 18 By the tangent-secant theorem, when a tangent and secant are drawn from the same external point, $PT^2 = PA \cdot PB$ . Substituting the known values: $12^2 = 8 \cdot PB$ , so $144 = 8 \cdot PB$ , giving $PB = 18$ . Choice A (16) might result from incorrectly using $PT \cdot PA = PB$ . Choice C (20) could come from using $PT + PA = PB$ . Choice D (24) might result from using $2 \cdot PT = PB$ .
In circle $M$ , radius $MR$ is perpendicular to chord $ST$ at point $H$ . If the radius of the circle is 13 and $MH = 5$ , what is the length of chord $ST$ ?
Answer: 24 When a radius is perpendicular to a chord, it bisects the chord. In right triangle $MHS$ (where $S$ is one endpoint of the chord), we have $MS = 13$ (radius), $MH = 5$ , and $HS$ can be found using the Pythagorean theorem: $HS^2 + MH^2 = MS^2$ , so $HS^2 + 25 = 169$ , giving $HS^2 = 144$ and $HS = 12$ . Since the radius bisects the chord, $ST = 2 \cdot HS = 2 \cdot 12 = 24$ . Choice A (12) is just half the chord length. Choice B (18) might result from adding $MH + MS = 5 + 13 = 18$ . Choice D (26) might come from adding $2 \cdot MS = 2 \cdot 13 = 26$ .
In the figure shown, $\angle APB$ is a central angle measuring $80¬∞$ , and $\angle AQB$ is an inscribed angle that intercepts the same arc $AB$ . Point $Q$ is then moved to a new position $Q'$ on the circle such that $\angle AQ'B$ intercepts the major arc $AB$ . What is the measure of $\angle AQ'B$ ?
Answer: 140¬∞ Initially, inscribed angle $\angle AQB$ intercepts the minor arc $AB$ and measures half the central angle: $\angle AQB = \frac{80¬∞}{2} = 40¬∞$ . When $Q$ moves to $Q'$ such that $\angle AQ'B$ intercepts the major arc $AB$ , this inscribed angle measures half of the major arc. The major arc $AB$ measures $360¬∞ - 80¬∞ = 280¬∞$ , so $\angle AQ'B = \frac{280¬∞}{2} = 140¬∞$ . Choice A (40¬∞) is the measure of the original inscribed angle. Choice B (100¬∞) might result from incorrectly adding $80¬∞ + 20¬∞$ . Choice D (280¬∞) is the arc measure, not the inscribed angle measure.
An inscribed angle and a central angle both intercept the same arc of a circle. If the central angle measures $3x + 20°$ and the inscribed angle measures $2x - 5°$ , what is the value of $x$ ?
Answer: 30 An inscribed angle measures half of the central angle that intercepts the same arc. Therefore: $2x - 5 = \frac{1}{2}(3x + 20)$ . Solving: $2x - 5 = \frac{3x + 20}{2}$ , so $2(2x - 5) = 3x + 20$ , giving $4x - 10 = 3x + 20$ , and $x = 30$ . Choice A (15) might result from setting the angles equal: $2x - 5 = 3x + 20$ . Choice B (25) could come from solving $2x = 3x + 20 - 5$ . Choice D (35) might result from incorrectly manipulating the equation.
Refer to the figure below. Two tangent lines from external point $P$ touch circle $O$ at points $A$ and $B$ . If $\angle APB = 46°$ , what is the measure of the minor arc $AB$ ?
Answer: $134°$ Radii $\overline{OA}$ and $\overline{OB}$ are perpendicular to the tangents, so quadrilateral $OAPB$ has two right angles. The remaining two angles sum to $180°$ , giving central angle $\angle AOB = 180° - 46° = 134°$ , which equals the minor arc $AB$ . Distractor A copies the given angle. B is half the correct central angle. D is the major arc.
Two chords $AB$ and $CD$ intersect inside circle $O$ at point $E$ . If $AE = 6$ , $EB = 4$ , and $CE = 8$ , what is the length of $ED$ ?
Answer: 3 When two chords intersect inside a circle, the products of their segments are equal: $AE \cdot EB = CE \cdot ED$ . Substituting: $6 \cdot 4 = 8 \cdot ED$ , so $24 = 8 \cdot ED$ , giving $ED = 3$ . Choice B (2) might result from using $AE - EB = ED$ . Choice C (5) could come from incorrectly using $\frac{AE + EB}{2} - 1 = ED$ . Choice D (12) might result from using $CE + EB = ED$ .

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