math-2017:G.C.2
math-2017:G.C.2 Identify and describe relationships among angles, radii, chords, tangents, and arcs and use them to solve problems. Include the relationship between central, inscribed, and circumscribed angles and their intercepted arcs; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.
Learning goal: Identify and describe relationships among angles, radii, chords, tangents and arcs of a circle and use them to solve problems: the relationship between central, inscribed and circumscribed angles and their intercepted arcs; why an inscribed angle on a diameter is a right angle; and why the radius of a circle is perpendicular to the tangent at the point where the radius meets the circle.
The vocabulary, with conditions. In a circle with center O: a radius is a segment from O to a point on the circle, and all radii of one circle are congruent. A chord is a segment whose two endpoints are on the circle; a diameter is a chord through the center, and its length is twice the radius. An arc is a connected part of the circle; a central angle is an angle whose vertex is O and whose sides are radii, and the measure of the arc it cuts off (its intercepted arc) is defined to equal the measure of the central angle, so a full circle is 360°. An inscribed angle has its vertex on the circle and its sides are chords. A tangent is a line that meets the circle in exactly one point, the point of tangency. A circumscribed angle has its vertex outside the circle and its sides are two tangents to the circle.
Central angle equals its arc. This is a definition, not a theorem: if central angle AOB measures 100°, then arc AB (the minor arc, the one inside the angle) measures 100°, and the major arc, the rest of the circle, measures 360° − 100° = 260°.
The inscribed angle theorem. An inscribed angle measures half the arc it intercepts, which is the same as half the central angle on that arc. Proof in the basic case, where one side of the inscribed angle is a diameter: let angle ACB be inscribed with C on the circle and CB passing through the center O. Draw radius OA. Triangle OAC is isosceles because OA and OC are both radii, so its base angles are equal; call each x. The central angle AOB is an exterior angle of triangle OAC at O, so it equals the sum of the two remote interior angles: angle AOB = x + x = 2x. Hence the inscribed angle x is half the central angle, and therefore half the arc AB. The other two cases, where the center lies inside the angle or outside it, follow by drawing the diameter from C and adding or subtracting two basic-case angles.
Consequences worth stating. Every inscribed angle intercepting the same arc has the same measure, since each is half that arc. An inscribed angle that intercepts a semicircle intercepts an arc of 180°, so it measures 90°: an inscribed angle on a diameter is a right angle. Conversely, if an inscribed angle is a right angle, its intercepted arc is 180°, so the chord joining its endpoints is a diameter. In a worked instance, if AB is a diameter of length 10 and C is another point on the circle with AC = 6, then triangle ACB is right-angled at C, and BC = √(10² − 6²) = √64 = 8.
A worked model inside the lesson. In a circle with center O, the central angle AOB measures 100° and C is a point on the major arc. Then arc AB = 100° (definition), inscribed angle ACB = ½·100° = 50° (inscribed angle theorem). Now let D be another point on the major arc: angle ADB is also 50°, because it intercepts the same arc. If instead E lies on the minor arc AB, then angle AEB intercepts the major arc, 260°, so angle AEB = 130°. Notice that 50° + 130° = 180°; this is the cyclic-quadrilateral fact proved in the next lesson.
Tangent and radius. The tangent line at a point T is perpendicular to the radius OT. Reason: every point of the tangent other than T lies outside the circle (a point of the line inside the circle would force the line to cross the circle a second time), so its distance from O is greater than r = OT. That makes T the point of the line closest to O, and the shortest segment from a point to a line is the perpendicular one; so OT is perpendicular to the tangent. The converse is also true: a line through a point T of the circle that is perpendicular to OT is the tangent at T. Because of the right angle, a tangent segment from an external point P forms a right triangle with the radius: if OP = 17 and the radius is 8, the tangent length PT satisfies PT² = 17² − 8² = 289 − 64 = 225, so PT = 15.
Two tangents from one external point. If PT and PS are tangent segments from the same external point P, then PT = PS. Both triangles OTP and OSP are right triangles (tangent ⟂ radius) sharing the hypotenuse OP and with congruent legs OT = OS, so they are congruent by hypotenuse-leg, and PT = PS.
Circumscribed angle and its arc. Let the tangents from P touch the circle at A and B. Quadrilateral PAOB has right angles at A and B, so its other two angles add to 360° − 180° = 180°: the circumscribed angle at P and the central angle AOB are supplementary. Since the central angle equals the minor arc AB, the circumscribed angle is 180° minus the minor arc. Equivalently it is half the difference of the major and minor arcs: with minor arc 130°, the major arc is 230°, the central angle is 130°, the circumscribed angle is 180° − 130° = 50°, and ½(230° − 130°) = 50° agrees.
Chords. A radius (or diameter) perpendicular to a chord bisects the chord and its arc, because the two right triangles formed with the endpoints share a leg and have congruent hypotenuses (radii). So if a chord of a circle with radius 13 is at distance 5 from the center, half the chord is √(13² − 5²) = √144 = 12 and the chord is 24. Congruent chords cut off congruent arcs and are equidistant from the center, and conversely; both follow from congruent isosceles triangles with radii as legs.
The misconception: an inscribed angle equals its arc. Students who remember "angle equals arc" for central angles apply it to inscribed angles too, doubling the true value. The vertex location decides the rule: vertex at the center, angle equals arc; vertex on the circle, angle is half the arc; vertex outside on two tangents, angle is 180° minus the minor arc. Always locate the vertex first.
Worked inscribed angle chain: In circle O, arc AB measures 84°. Find the central angle AOB, the inscribed angle ACB for C on the major arc, and the inscribed angle AEB for E on the minor arc. Central angle: 84°, by definition of arc measure. Angle ACB: half the intercepted arc, ½·84° = 42°. Angle AEB intercepts the major arc, 360° − 84° = 276°, so angle AEB = 138°. Check: 42° + 138° = 180°, as opposite angles of the inscribed quadrilateral ACBE must be.
Worked right angle on a diameter: AB is a diameter of a circle of radius 6.5, and C is on the circle with BC = 5. Because angle ACB is inscribed on a diameter, it is 90°, so triangle ACB is a right triangle with hypotenuse AB = 13. Then AC = √(13² − 5²) = √(169 − 25) = √144 = 12. The area of the triangle is ½·5·12 = 30 square units.
Worked tangent from an external point: Point P is 26 units from the center O of a circle of radius 10, and PT is tangent at T. The radius OT is perpendicular to the tangent, so triangle OTP has a right angle at T with hypotenuse OP. PT² = 26² − 10² = 676 − 100 = 576, so PT = 24. The second tangent segment PS from P also has length 24, because the two right triangles OTP and OSP are congruent by hypotenuse-leg.
Worked circumscribed angle: Tangents from P touch circle O at A and B, and angle APB = 64°. In quadrilateral PAOB the angles at A and B are right angles, so angle AOB = 360° − 90° − 90° − 64° = 116°. The minor arc AB therefore measures 116° and the major arc measures 244°. Check by the half-difference form: ½(244° − 116°) = ½·128° = 64°.
Worked chord: A chord of length 30 lies in a circle of radius 17. The perpendicular from the center bisects the chord into two segments of 15, forming a right triangle with hypotenuse 17. The distance from the center to the chord is √(17² − 15²) = √(289 − 225) = √64 = 8.
Vertex-location table: Draw one circle and mark a central angle, an inscribed angle, a circumscribed angle and an inscribed angle on a diameter, each intercepting an arc you label with a measure of your own choosing (use different measures for each). Under the drawing make a four-row table with columns: where the vertex is, the rule relating angle to arc, the arc measure, the angle measure. Add a fifth row for a tangent and the radius at its point of tangency, giving the angle between them and the reason.
Independent performance: Design a problem of your own about a circle in which a diameter, a chord, a tangent and an inscribed angle all appear, with at least three numerical facts given (for example a radius, an arc measure and one segment length), and solve it completely. Success criteria: each step names the relationship it uses (definition of arc measure, inscribed angle theorem, right angle on a diameter, tangent perpendicular to radius, perpendicular from center bisects chord); every length found is checked with the Pythagorean Theorem or by a second route; every angle found is consistent with the 360° total of the circle; and one sentence explains why an inscribed angle is half, not equal to, its intercepted arc.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
Central angle AOB measures 118°. Find the measure of minor arc AB and of an inscribed angle ACB with C on the major arc. State the rule for each.
Minor arc AB = 118°, because the measure of an arc is defined as the measure of its central angle. Angle ACB = ½·118° = 59°, because an inscribed angle measures half its intercepted arc.
Prove that an inscribed angle is half the central angle on the same arc, in the case where one side of the inscribed angle passes through the center.
Let angle ACB be inscribed with CB through the center O, and draw radius OA. OA = OC (radii), so triangle OAC is isosceles and angles OAC and OCA are equal; call each x. Angle AOB is the exterior angle of triangle OAC at O, so it equals x + x = 2x. Hence angle ACB = x = ½·angle AOB = ½·arc AB.
AB is a diameter of a circle and C is a point on the circle. If AC = 9 and BC = 12, find the radius and explain why the triangle is right-angled.
Angle ACB is inscribed on a diameter, so it intercepts a 180° arc and measures 90°. Then AB² = 9² + 12² = 81 + 144 = 225, AB = 15, and the radius is 7.5.
Explain why a tangent line is perpendicular to the radius drawn to the point of tangency.
Every point of the tangent other than the point of tangency T lies outside the circle, so it is farther from the center O than the radius OT. T is therefore the closest point of the line to O, and the shortest segment from a point to a line is perpendicular to the line. So OT ⟂ tangent.
Point P is 25 units from the center of a circle of radius 7. Find the length of a tangent segment from P and explain why the two tangent segments from P are equal.
The radius to the point of tangency is perpendicular to the tangent, so PT² = 25² − 7² = 625 − 49 = 576 and PT = 24. The two right triangles formed by the two tangent segments share hypotenuse OP and have congruent legs (radii), so they are congruent by hypotenuse-leg and the tangent segments are equal, both 24.
Two tangents from P touch circle O at A and B, and minor arc AB measures 140°. Find the circumscribed angle APB two ways.
Quadrilateral PAOB has right angles at A and B, so angle APB = 360° − 90° − 90° − 140° = 40°. Alternatively the major arc is 220°, and ½(220° − 140°) = 40°.
A chord of a circle of radius 10 is 6 units from the center. Find its length, name the fact that makes the computation possible, and explain why a chord can never be 12 units from the center of this circle.
The perpendicular from the center bisects the chord, so half the chord is a leg of a right triangle with hypotenuse 10 (a radius) and other leg 6. Half the chord is √(100 − 36) = √64 = 8, so the chord is 16. A chord 12 units away is impossible because the leg of a right triangle cannot exceed its hypotenuse: every point of a chord is at most 10 units from the center.
Points C and D both lie on the major arc AB, and angle ACB = 37°. Find angle ADB and the measure of minor arc AB, and explain.
Angle ADB = 37°, because inscribed angles that intercept the same arc are equal (each is half the arc). Minor arc AB = 2·37° = 74°, because the inscribed angle is half its intercepted arc.