Circles — Free Geometry Review Games.

A central angle equals its intercepted arc.

A diameter creates two equal semicircles (180 degrees each).

A tangent is always perpendicular to the radius drawn to the point of tangency.

An inscribed angle is half the measure of its intercepted arc.

A chord through the center is the diameter.

Inscribed angle = half the arc = 100/2 = 50 degrees.

Arc length = (120/360)*2*pi*6 = (1/3)*12*3.14 = 12.56.

Tangent segments from the same external point are congruent.

A semicircle is 180 degrees, so the inscribed angle = 180/2 = 90 degrees.

Intersecting chords theorem: the products of the segments are equal.

3*8 = 4*x, 24 = 4x, x = 6.

tangent^2 = external*whole: 36 = 4*whole, whole = 9.

Area = (60/360)*pi*81 = (1/6)*254.34 = 42.39.

Inscribed angle = half the arc = 200/2 = 100 degrees.

Angle = (far arc - near arc)/2 = (130-70)/2 = 30. Correct.

A complete circle measures \(360^\circ\). This is the foundation of the degree system used to measure arcs and central angles. A common distractor is \(180^\circ\), which is the measure of a semicircle (half a circle), not a full circle.

A minor arc spans less than half of a circle, so its degree measure is strictly less than \(180^\circ\). An arc measuring exactly \(180^\circ\) is a semicircle, not a minor arc. An arc greater than \(180^\circ\) is called a major arc.

The Equidistant Chords Theorem states that two chords equidistant from the center of a circle are congruent. The perpendicular distance from the center determines chord length — equal distances yield equal chord lengths. 'Parallel' is a common misconception: equal distance from the center does not require a parallel orientation.

The Perpendicular Bisector Theorem for circles states that a perpendicular from the center to a chord bisects the chord, dividing it into two equal segments. It also bisects the corresponding arc, so choice B is incorrect because it describes only part of the full result.

A tangent line touches a circle at exactly one point, called the point of tangency. A line that intersects a circle at two points is a secant, not a tangent. A line with zero intersection points lies entirely outside the circle.

A diameter passes through the center and has endpoints on the circle, giving it the maximum possible length of \(2r\). A radius is only half as long. A tangent segment is not a chord at all because it only touches the circle at one point rather than having both endpoints on the circle.

A major arc spans more than half the circle, so its measure must be greater than \(180^\circ\). An arc measuring exactly \(180^\circ\) is a semicircle. Any arc with measure less than \(180^\circ\) is a minor arc, regardless of whether it is less than or greater than \(90^\circ\).

The Tangent-Chord Angle Theorem states that the angle between a tangent and a chord equals half the intercepted arc. So \(50^\circ = \frac{1}{2} \cdot \text{arc}\), which gives \(\text{arc} = 100^\circ\). A common error is choosing \(50^\circ\), which confuses the angle with the arc — they are only equal when the arc measures \(180^\circ\) (a semicircle).

Inscribed angles that intercept the same arc are congruent because each equals half the intercepted arc. If both angles intercept arc \(\widehat{AB}\) with measure \(m\), then each angle equals \(\frac{1}{2}m\), making them equal. They are not supplementary — supplementary angles sum to \(180^\circ\), which is not guaranteed here.

When two chords intersect inside a circle, the angle equals half the sum of the two intercepted arcs: \(\angle = \frac{1}{2}(80^\circ + 60^\circ) = \frac{1}{2}(140^\circ) = 70^\circ\). Choosing \(80^\circ\) is a common error — that uses only one arc without averaging. Choosing \(40^\circ\) results from taking half of just the \(80^\circ\) arc.

Arc length is given by \(s = \frac{\theta}{360^\circ} \cdot 2\pi r = \frac{60}{360} \cdot 2\pi(6) = \frac{1}{6} \cdot 12\pi = 2\pi\). A common error is using the sector area formula \(\frac{\theta}{360^\circ} \cdot \pi r^2\) instead, which measures area rather than length.

For an angle formed outside the circle by a secant and a tangent, the angle equals half the positive difference of the intercepted arcs: \(\angle = \frac{1}{2}(180^\circ - 60^\circ) = \frac{1}{2}(120^\circ) = 60^\circ\). Choosing \(120^\circ\) is a frequent error — that is the raw difference before dividing by \(2\).

The perpendicular from the center bisects the chord, forming a right triangle with hypotenuse \(r = 10\) and one leg equal to half the chord \(= 6\). By the Pythagorean theorem: \(d = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\) cm. Choosing \(6\) cm is a typical mistake — that is half the chord length, not the distance from the center.

The Inscribed Angle Theorem states that a central angle is twice the inscribed angle intercepting the same arc: $\angle BOC = 2 \times \angle BAC = 2 \times 42^\circ = 84^\circ$. Choosing \(42^\circ\) is the most common error — students confuse the inscribed angle with the central angle, but they are only equal when both measure \(90^\circ\).

In a cyclic quadrilateral, opposite angles are supplementary and sum to \(180^\circ\). The angle opposite the \(80^\circ\) angle is \(180^\circ - 80^\circ = 100^\circ\). Verification: \(65^\circ + 115^\circ = 180^\circ\) and \(80^\circ + 100^\circ = 180^\circ\), with all four summing to \(360^\circ\). Choosing \(80^\circ\) incorrectly assumes opposite angles are equal, a property of parallelograms, not cyclic quadrilaterals in general.

The perpendicular from the center bisects the chord. Using the Pythagorean theorem, half the chord length \(= \sqrt{r^2 - d^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8\). The total chord length is \(2 \times 8 = 16\) units. Choosing \(8\) units is the most common error — it represents only half the chord.

Using the arc length formula \(s = \frac{\theta}{360^\circ} \cdot 2\pi r\): \(8\pi = \frac{\theta}{360} \cdot 24\pi\), so \(\frac{\theta}{360} = \frac{1}{3}\), giving \(\theta = 120^\circ\). Equivalently in radians, \(\theta = \frac{s}{r} = \frac{8\pi}{12} = \frac{2\pi}{3}\) radians \(= 120^\circ\).

The minor arc and major arc together form the complete circle: \(\text{minor arc} + \text{major arc} = 360^\circ\). So major arc \(= 360^\circ - 100^\circ = 260^\circ\). Choosing \(180^\circ\) is a common misconception — the two arcs only sum to \(180^\circ\) each when the chord connecting the tangent points is a diameter.

By the Power of a Point theorem for two secants from an external point: \((\text{external}_1)(\text{whole}_1) = (\text{external}_2)(\text{whole}_2)\). So \(3 \times 12 = 4 \times x\), giving \(36 = 4x\) and \(x = 9\). Choosing \(8\) is a common error from writing \(3 + 12 = 4 + x\) (adding instead of multiplying), which misapplies the theorem.

In a cyclic quadrilateral, opposite angles are supplementary: \(\angle A + \angle C = 180^\circ\). Setting up the equation: \((2x + 15) + (x + 45) = 180\), so \(3x + 60 = 180\), giving \(3x = 120\) and \(x = 40\). Substituting back: \(\angle A = 95^\circ\) and \(\angle C = 85^\circ\), which correctly sum to \(180^\circ\). Checking \(x = 35\) gives \(85^\circ + 80^\circ = 165^\circ \neq 180^\circ\), confirming it is wrong.

When two chords are parallel, the arcs between the parallel chords on each side are congruent: arc \(\widehat{AC} =\) arc \(\widehat{BD} = 70^\circ\). Since the total circle is \(360^\circ\): arc \(\widehat{AB} = 360^\circ - 70^\circ - 100^\circ - 70^\circ = 120^\circ\). Choosing \(100^\circ\) is incorrect — it assumes arc \(\widehat{AB}\) equals arc \(\widehat{CD}\), which would only hold if the chords were also congruent in a specific symmetric configuration.

The angle formed by two tangents from an external point equals half the positive difference of the intercepted arcs. Let the minor arc \(= m\), so the major arc \(= 360^\circ - m\). Then \(40^\circ = \frac{1}{2}\left[(360^\circ - m) - m\right] = \frac{1}{2}(360^\circ - 2m) = 180^\circ - m\). Solving: \(m = 180^\circ - 40^\circ = 140^\circ\). Choosing \(80^\circ\) is a common error from simply doubling the angle (\(2 \times 40^\circ\)) without applying the correct formula.

A radius drawn to a point of tangency is perpendicular to the tangent, forming a right triangle with the radius as one leg, the tangent as the other leg, and \(OP\) as the hypotenuse. By the Pythagorean theorem: \(PT^2 = OP^2 - r^2 = 13^2 - 5^2 = 169 - 25 = 144\), so \(PT = 12\). Choosing \(18\) comes from incorrectly adding \(r + OP = 5 + 13\), which has no geometric justification.

By the Intersecting Chords theorem: \(AP \cdot PB = CP \cdot PD\), so \(3 \times 12 = 4 \times PD\), giving \(36 = 4 \cdot PD\) and \(PD = 9\). The total length of chord \(\overline{CD} = CP + PD = 4 + 9 = 13\). Choosing \(9\) is a common error — that is only segment \(PD\), not the full chord length.

The major arc \(\widehat{AB} = 360^\circ - 120^\circ = 240^\circ\). The angle formed by two tangents from an external point equals half the positive difference of the intercepted arcs: $\angle APB = \frac{1}{2}(240^\circ - 120^\circ) = \frac{1}{2}(120^\circ) = 60^\circ$. Choosing \(120^\circ\) confuses the angle with the minor arc. Choosing \(30^\circ\) results from halving the minor arc directly without first computing the arc difference.

A central angle and the arc it intercepts always have equal measure, so \(\widehat{AB}\) must be \(100^\circ\). The choice \(50^\circ\) confuses this rule with the inscribed angle theorem, where the inscribed angle (not the central angle) is half the arc. Remember: central angle = intercepted arc, but inscribed angle = half the intercepted arc.

The Inscribed Angle Theorem states that an inscribed angle measures exactly half of the central angle subtending the same arc, since the inscribed angle's vertex lies on the circle rather than the center. The choice 'Twice' reverses the relationship and would make the inscribed angle larger than the central angle, which is impossible. This half-relationship is the foundation for solving almost every inscribed angle problem on the exam.

A tangent line is always perpendicular to the radius drawn to the point of tangency, so the angle formed is \(90^\circ\) by the Tangent-Radius Theorem. The choice \(180^\circ\) would mean the radius and tangent line are collinear, which contradicts the definition of tangency. This perpendicularity is frequently the key first step in tangent-length and tangent-circle problems.

By Thales' Theorem, any inscribed angle that intercepts a semicircle (a diameter) is a right angle, so $m\angle ABC = 90^\circ$. The choice \(180^\circ\) mistakes the arc measure of the semicircle for the inscribed angle, which is actually half of that arc. Whenever a triangle inscribed in a circle has a side that is a diameter, the angle opposite that side is automatically \(90^\circ\).

All central angles around the center of a circle must sum to \(360^\circ\), so \(x = 360 - 110 - 95 = 155^\circ\). The choice \(145^\circ\) results from an arithmetic slip in the subtraction and does not satisfy the total-degree requirement. Always remember that central angles surrounding a single point sum to a full \(360^\circ\) rotation.

A tangent line touches a circle at exactly one point, called the point of tangency, without crossing into the interior. A 'secant' is incorrect because a secant line intersects the circle at two points, passing through its interior. Distinguishing tangents (one intersection) from secants and chords (two intersections) is essential vocabulary for circle theorems.

Arc length is found using \(\text{arc length} = 2\pi r \cdot \frac{\theta}{360}\), so \(2\pi(9)\cdot\frac{120}{360} = 6\pi\). The choice \(3\pi\) would result from using \(60^\circ\) instead of \(120^\circ\) in the fraction. Always convert the central angle to a fraction of \(360^\circ\) before multiplying by the circumference.

The Tangent-Chord Angle Theorem states that the angle formed by a tangent and a chord equals half the intercepted arc, so $\angle PAB = \frac{130}{2} = 65^\circ$. The choice \(130^\circ\) mistakenly uses the full arc measure instead of taking half of it. This half-arc relationship mirrors the inscribed angle theorem and applies whenever a tangent meets a chord at the point of tangency.

When two secants meet outside a circle, the angle formed equals half the positive difference of the intercepted arcs, so \(\angle P = \frac{150-50}{2} = 50^\circ\). The choice \(100^\circ\) comes from forgetting to divide the difference of arcs by two. This 'half the difference' rule applies to any angle whose vertex lies outside the circle, whether formed by two secants, two tangents, or a secant and a tangent.

When two chords intersect inside a circle, the angle formed equals half the sum of the two intercepted arcs, so $\angle AEC = \frac{80+40}{2} = 60^\circ$. The choice \(120^\circ\) uses the full sum of the arcs without dividing by two. This 'half the sum' rule for interior intersections is the mirror image of the 'half the difference' rule used for exterior intersection points.

Inscribed angles that intercept the same arc are always congruent, since both equal half the same arc measure by the Inscribed Angle Theorem, so $m\angle ADC = 37^\circ$. The choice \(74^\circ\) incorrectly doubles the angle, as if computing the arc rather than the second inscribed angle. Whenever two inscribed angles share the same intercepted arc, they must be equal regardless of where their vertices sit on the circle.

Since the three arcs must sum to \(360^\circ\), \(9x = 360\) gives \(x = 40\), so arc \(\widehat{AB} = 3(40) = 120^\circ\), and the central angle equals this arc measure directly. The choice \(40^\circ\) mistakenly reports the value of \(x\) rather than the actual central angle. Whenever arcs are given in terms of a variable, always solve for the variable using the 360-degree total before finding the requested angle.

By the Tangent-Secant Power of a Point relationship, \(\text{tangent}^2 = \text{external} \times \text{whole secant}\), so \(8^2 = 4 \times (4+x)\), giving \(64 = 16 + 4x\) and \(x = 12\). The choice \(16\) represents the whole secant length rather than just the far segment beyond the near intersection point. Always be careful to distinguish the 'external segment,' the 'far segment,' and the 'whole secant' when applying this theorem.

Using \(\text{arc length} = 2\pi r \cdot \frac{\theta}{360}\), we set \(5\pi = 2\pi r \cdot \frac{90}{360} = \frac{\pi r}{2}\), so solving gives \(r = 10\). The choice \(5\) would result from forgetting to account for the fraction of the circle represented by \(90^\circ\). When working backward from arc length to radius, always isolate \(r\) carefully after substituting the given central angle fraction.

The angle formed by two tangents from an external point equals half the difference of the intercepted arcs, so with major arc \(\widehat{AB} = 360-80=280^\circ\), \(m\angle P = \frac{280-80}{2} = 100^\circ\). The choice \(80^\circ\) mistakenly uses the minor arc directly as the angle instead of applying the half-difference formula. Whenever tangents meet outside a circle, remember to find the major arc first before applying the exterior angle formula.

Using the exterior angle formula \(m\angle P = \frac{\text{far arc} - \text{near arc}}{2}\), we get \(25 = \frac{4x-(2x-10)}{2}\), which simplifies to \(50 = 2x+10\), so \(x=20\) and the far arc equals \(4(20)=80^\circ\). The choice \(30^\circ\) is the value of the near arc, not the far arc that the question requests. When solving these algebraic secant-angle problems, always verify which arc — near or far — the question is actually asking for.

By the Tangent-Chord Angle Theorem, the angle equals half the intercepted arc, so \(2(x+10) = 3x-20\), giving \(2x+20=3x-20\) and \(x=40\), so the arc equals \(3(40)-20=100^\circ\). The choice \(50^\circ\) is actually the value of the tangent-chord angle itself, not the arc the question asks for. Setting up the equation as angle \(=\) half the arc (or doubling the angle to equal the arc) is the key algebraic step in these problems.

By the Intersecting Chords Theorem, \(AE \times EB = CE \times ED\), so \(6 \times 8 = x(x+2)\), giving \(x^2+2x-48=0\), which factors to \((x+8)(x-6)=0\), so \(x=6\) (the negative root is rejected since lengths must be positive). The choice \(8\) satisfies neither the factored equation nor the product requirement of \(48\). Whenever two chords intersect, remember that the products of their respective segments must always be equal.

Since a tangent segment is perpendicular to the radius at the point of tangency, triangle $OAP$ is a right triangle with legs \(OA=5\) and \(PA\), and hypotenuse \(OP=13\), so by the Pythagorean Theorem \(PA=\sqrt{13^2-5^2}=\sqrt{144}=12\). The choice \(10\) does not satisfy the Pythagorean relationship among the given values of \(5\) and \(13\). Whenever a tangent segment and a radius meet, drawing the right triangle formed with the segment to the external point is often the fastest path to a solution.

Since the three arcs sum to \(360^\circ\), \(5x+3x+20+4x-20=12x=360\), so \(x=30\), making arc \(\widehat{CA}=4(30)-20=100^\circ\); by the Inscribed Angle Theorem, $\angle ABC$ equals half of the arc it intercepts, which is arc \(\widehat{CA}\), so $\angle ABC = 50^\circ$. The choice \(100^\circ\) mistakenly reports the arc measure itself instead of taking half of it for the inscribed angle. This problem shows the importance of first solving for the variable using the arc-sum property, then correctly identifying which arc the requested inscribed angle actually intercepts.