Circle Angles & Arcs - Math 2

In circle $O$ , secants $PAB$ and $PCD$ are drawn from external point $P$ , where $A$ and $C$ are the closer intersection points to $P$ . If $\angle APD = 40°$ and arc $BD = 140°$ , what is the measure of arc $AC$ ?
Answer: $60°$ When two secants are drawn from an external point, the angle between them equals half the difference of the intercepted arcs. Here, $\angle APD = \frac{1}{2}(\text{arc } BD - \text{arc } AC)$ . Substituting the known values: $40° = \frac{1}{2}(140° - \text{arc } AC)$ . Solving: $80° = 140° - \text{arc } AC$ , so $\text{arc } AC = 140° - 80° = 60°$ . Choice B ( $80°$ ) comes from incorrectly using $2 \times 40° = 80°$ . Choice C ( $100°$ ) might result from $140° - 40°$ . Choice D ( $220°$ ) could come from using $140° + 80°$ .
A tangent from external point $T$ touches circle $O$ at point $A$ . A secant from $T$ passes through points $B$ and $C$ on the circle (with $B$ closer to $T$ ). If arc $AC = 140°$ and the angle between the tangent and secant is $35°$ , what is the measure of arc $AB$ ?
Answer: $70°$ When you see a tangent and secant drawn from an external point, you're dealing with the angle-arc relationship for angles formed outside a circle. The key theorem states that an angle formed by two secants, two tangents, or a tangent and secant from an external point equals half the difference of the intercepted arcs. Here, the angle between tangent $TA$ and secant $TBC$ is $35°$ . The intercepted arcs are arc $AC = 140°$ (given) and arc $AB$ (what we're finding). Using the theorem: $35° = \frac{1}{2}|$ arc $AC$ - arc $AB$ |$$ $35° = \frac{1}{2}|140° -$ arc $AB$ |$$ $70° = |140° -$ arc $AB$ | $This gives us two cases:$ 140° - $arc$ AB = 70° $or$ 140° - $arc$ AB = -70° $Solving: arc$ AB = 70° $or arc$ AB = 210° $Since$ B $is closer to$ T $than$ C $, and both points lie on the circle with the tangent at$ A $, the geometric configuration requires arc$ AB = 70° $. Looking at the wrong answers: (A)$ 210° $is the other mathematical solution but doesn't fit the geometric constraint. (B)$ 105° $likely comes from incorrectly using$ 35° × 3 $. (C)$ 175° $might result from adding$ 35° $to$ 140°$$ instead of using the proper theorem. Remember: Always apply the geometric constraints after solving algebraically. The position of points matters in circle problems, not just the arithmetic.
Circle $R$ has inscribed quadrilateral $WXYZ$ . If arc $WX = 90°$ , arc $XY = 70°$ , and arc $YZ = 110°$ , what is the measure of $\angle WXY + \angle WZY$ ?
Answer: $180°$ In an inscribed quadrilateral, opposite angles are supplementary. $\angle WXY$ and $\angle WZY$ are opposite angles, so $\angle WXY + \angle WZY = 180°$ . This can also be verified by calculating: arc $ZW = 360° - 90° - 70° - 110° = 90°$ . $\angle WXY$ intercepts arc $WZY = 110° + 90° = 200°$ , so $\angle WXY = 100°$ . $\angle WZY$ intercepts arc $WXY = 90° + 70° = 160°$ , so $\angle WZY = 80°$ . Thus $100° + 80° = 180°$ . Choice A uses one arc measure. Choice B is the sum of two adjacent arc measures. Choice D uses three-quarters of the circle.
In circle $O$ , central angle $\angle AOB = 72°$ and inscribed angle $\angle ACB$ intercepts the same arc $AB$ . Point $D$ is on the circle such that inscribed angle $\angle ADB$ intercepts arc $AB$ from the opposite side of the circle. What is the measure of $\angle ACB + \angle ADB$ ?
Answer: $72°$ Since central angle $\angle AOB = 72°$ , the arc $AB$ measures $72°$ . An inscribed angle measures half the intercepted arc, so $\angle ACB = 36°$ . Since $\angle ADB$ also intercepts arc $AB$ (though from the opposite side), it also measures $36°$ . Therefore, $\angle ACB + \angle ADB = 36° + 36° = 72°$ . Choice B incorrectly adds the central angle to one inscribed angle. Choice C doubles the central angle. Choice D incorrectly uses the reflex arc measure.
Circle $M$ has center $(3, 4)$ and radius $5$ . Points $A(-2, 4)$ and $B(8, 4)$ are on the circle. Point $C$ is also on the circle such that triangle $ABC$ is inscribed in the circle. If $C$ is in the upper half of the circle, what is the measure of $\angle ACB$ ?
Answer: $90°$ First, verify that $A$ and $B$ are on the circle: distance from $M(3,4)$ to $A(-2,4)$ is $\sqrt{(-2-3)^2 + (4-4)^2} = 5$ ✓. Distance from $M$ to $B(8,4)$ is $\sqrt{(8-3)^2 + (4-4)^2} = 5$ ✓. Since $A(-2,4)$ and $B(8,4)$ have the same $y$ -coordinate and the center $M(3,4)$ also has $y = 4$ , the chord $AB$ passes through the center, making it a diameter. Any inscribed angle that intercepts a diameter (semicircle) measures $90°$ . Therefore, $\angle ACB = 90°$ . Choice A would be for a $90°$ arc. Choice B would be for a $120°$ arc. Choice D would be for a $240°$ arc.
In circle $O$ , inscribed quadrilateral $ABCD$ has $\angle A = 85°$ and $\angle B = 110°$ . A diagonal $AC$ divides the quadrilateral into two triangles. What is the difference between the measures of $\angle C$ and $\angle D$ ?
Answer: $25°$ In an inscribed quadrilateral, opposite angles are supplementary. So $\angle A + \angle C = 180°$ and $\angle B + \angle D = 180°$ . Given $\angle A = 85°$ , we have $\angle C = 180° - 85° = 95°$ . Given $\angle B = 110°$ , we have $\angle D = 180° - 110° = 70°$ . Therefore, $\angle C - \angle D = 95° - 70° = 25°$ . Choice A suggests they're equal. Choice C is the value of $\angle D$ . Choice D is the sum of $\angle A$ and $\angle B$ minus some value.
Circle $P$ has center at the origin. Points $A(3, 4)$ , $B(-4, 3)$ , and $C$ are on the circle. If inscribed angle $\angle BAC = 30°$ , what is the measure of the minor arc $BC$ ?
Answer: $60°$ An inscribed angle measures half the intercepted arc. Since $\angle BAC = 30°$ , the arc $BC$ that it intercepts measures $2 \times 30° = 60°$ . Choice A incorrectly equates the inscribed angle with the arc. Choice C might result from confusing this with a right angle relationship. Choice D doubles the correct answer, possibly confusing the relationship direction.
A regular hexagon is inscribed in a circle. If one side of the hexagon subtends an arc of measure $x°$ , and an inscribed angle is drawn from a vertex of the hexagon to subtend three consecutive sides, what is the measure of this inscribed angle in terms of $x$ ?
Answer: $\frac{3x}{2}$ In a regular hexagon inscribed in a circle, each side subtends an equal arc. Since the total circle is $360°$ and there are 6 sides, each side subtends an arc of $\frac{360°}{6} = 60°$ . So $x = 60°$ . Three consecutive sides subtend a total arc of $3x$ . An inscribed angle that intercepts this arc has measure $\frac{3x}{2}$ (half the intercepted arc). Choice A ( $\frac{x}{2}$ ) would be the inscribed angle intercepting just one side. Choice C ( $3x$ ) incorrectly assumes the angle equals the arc. Choice D ( $\frac{3x}{4}$ ) has no geometric basis for this relationship.
In circle $M$ , chord $AB$ subtends a central angle of $100°$ . Point $C$ is on the major arc $AB$ , and point $D$ is on the minor arc $AB$ . What is $\angle ACB - \angle ADB$ ?
Answer: $-40°$ Since the central angle is $100°$ , the minor arc $AB = 100°$ and the major arc $AB = 360° - 100° = 260°$ . Point $C$ on the major arc creates inscribed angle $\angle ACB$ that intercepts the minor arc, so $\angle ACB = 100°/2 = 50°$ . Point $D$ on the minor arc creates inscribed angle $\angle ADB$ that intercepts the major arc, so $\angle ADB = 260°/2 = 130°$ . Therefore, $\angle ACB - \angle ADB = 50° - 130° = -40°$ . Choice C reverses the subtraction order. Choice B assumes they're equal. Choice D adds them instead.
Two chords $AB$ and $CD$ intersect inside circle $O$ at point $P$ . If arc $AC = 80°$ and arc $BD = 40°$ , what is the measure of $\angle APD$ ?
Answer: $60°$ When two chords intersect inside a circle, the measure of each angle formed equals half the sum of the intercepted arcs. $\angle APD$ intercepts arcs $AC$ and $BD$ , so $\angle APD = \frac{80° + 40°}{2} = 60°$ . Choice A uses only one arc divided by 2. Choice C uses just one arc. Choice D uses the sum without dividing by 2.
A regular hexagon is inscribed in circle $O$ . If one side of the hexagon subtends a central angle, and $P$ is a point on the circle not at any vertex of the hexagon, what is the measure of the inscribed angle $\angle APB$ where $A$ and $B$ are adjacent vertices of the hexagon?
Answer: $30°$ When you encounter problems involving regular polygons inscribed in circles, remember that inscribed angles and central angles have a fundamental relationship: an inscribed angle is always half the central angle that subtends the same arc. First, let's find the central angle. A regular hexagon has 6 equal sides, so each side subtends a central angle of $\frac{360°}{6} = 60°$ . This means the arc $AB$ (between adjacent vertices) measures $60°$ . Now, since $P$ is any point on the circle (not at a vertex), angle $\angle APB$ is an inscribed angle that intercepts the same arc $AB$ . By the inscribed angle theorem, this inscribed angle equals half the central angle: $\frac{60°}{2} = 30°$ . Looking at the wrong answers: Choice A ( $120°$ ) incorrectly doubles the central angle instead of halving it. Choice B ( $60°$ ) gives you the central angle itself—this is a common mistake when students forget to apply the inscribed angle theorem. Choice C ( $90°$ ) might seem tempting if you incorrectly think about the hexagon's internal angles, but those aren't relevant here. The correct answer is D ( $30°$ ). Key strategy: For any regular $n$ -sided polygon inscribed in a circle, each side subtends a central angle of $\frac{360°}{n}$ . Any inscribed angle intercepting that same arc will be exactly half that value. This pattern appears frequently on geometry exams, so memorizing the inscribed angle theorem will save you time.

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