6.2 Chords, Tangents & Secants
Key Takeaways
- A tangent line is perpendicular to the radius at the point of tangency, creating a usable right angle.
- Two tangent segments drawn from the same external point are congruent.
- A perpendicular from the center bisects a chord; chord length = 2 * sqrt(r squared - d squared).
- Angle rule by vertex: on the circle = half one arc, inside = half the SUM, outside = half the DIFFERENCE.
- Power of a point: chord-chord gives a*b = c*d; secant-tangent gives tangent squared = whole * external.
Tangents, Chords, and the Right Angles They Create
Once the two basic angle theorems from Section 6.1 are secure, the Regents adds relationships involving tangents, chords, and secants. These reward the same instinct: locate the vertex, then apply the matching rule. This section covers standard GEO-G.C.2.
Tangent-Radius Perpendicularity
The foundation for nearly every tangent problem is this: a tangent line is perpendicular to the radius drawn to the point of tangency. That guaranteed right angle lets you build a right triangle and use the Pythagorean theorem.
The same fact appears in coordinate geometry. The circle x squared + y squared = 25 passes through T(3, 4). The radius from the origin to T has slope 4/3, so the tangent at T is perpendicular to it and has the negative-reciprocal slope, -3/4.
The right angle is also the key to distance problems. If a tangent segment of length 8 is drawn from an external point P to a circle of radius 6, then the radius, the tangent, and the segment from P to the center form a right triangle with the right angle at the point of tangency. The distance from P to the center is sqrt(6 squared + 8 squared) = sqrt(100) = 10. Any time you see "tangent" and "external point" together, look for this radius-tangent right triangle.
Tangent Segments From an External Point
From a single external point, the two tangent segments to a circle are congruent. If tangents PA and PB are drawn from point P, then PA = PB. This turns into a quick equation on the Regents:
If PA = 3x + 1 and PB = x + 9, set them equal: 3x + 1 = x + 9, so 2x = 8 and x = 4. Then PA = 3(4) + 1 = 13.
Chord Properties
A perpendicular drawn from the center to a chord bisects that chord. This creates a right triangle whose hypotenuse is a radius, one leg is the distance from the center to the chord, and the other leg is half the chord. The chord-length shortcut is chord = 2 * sqrt(r squared - d squared), where d is the distance from center to chord.
Worked example: a chord lies 5 units from the center of a circle of radius 13. Half the chord is sqrt(13 squared - 5 squared) = sqrt(169 - 25) = sqrt(144) = 12, so the whole chord is 24.
Angle Rules: Where Is the Vertex?
The measure of an angle formed by chords, secants, or tangents depends entirely on where the vertex sits.
| Vertex location | Angle rule |
|---|---|
| On the circle (tangent-chord or two chords) | half of ONE intercepted arc |
| Inside the circle (two chords cross) | half the SUM of the two intercepted arcs |
| Outside the circle (two secants, secant-tangent, or two tangents) | half the DIFFERENCE of the two intercepted arcs |
A tangent-chord angle (vertex on the circle) equals half its intercepted arc. If that angle is 55 degrees, the intercepted arc is 2 * 55 = 110 degrees. Remember the pattern: on the circle uses one arc, inside uses the sum, outside uses the difference.
Segment Rules: Power of a Point
The same three vertex positions govern segment lengths through the power of a point relationships.
- Chord-chord (inside): when two chords cross, the products of their pieces are equal, a * b = c * d.
- Secant-secant (outside): for each secant, whole segment times external segment are equal, (whole1)(external1) = (whole2)(external2).
- Secant-tangent (outside): the tangent squared equals the whole secant times its external part, tangent squared = (whole secant)(external secant).
Worked example (chord-chord): two chords intersect inside a circle. One is divided into pieces 4 and 6; the other into pieces 3 and x. Then 4 * 6 = 3 * x, so 24 = 3x and x = 8.
Worked example (secant-tangent segment): a tangent segment of length 6 and a secant are drawn from the same external point. The secant's external part is 4 and its whole length is s. The tangent-squared rule gives 6 squared = 4 * s, so 36 = 4s and s = 9; the internal chord along the secant is therefore 9 - 4 = 5.
Worked example (outside angle): two secants from an external point intercept a far arc of 80 degrees and a near arc of 30 degrees. Because the vertex is outside the circle, the angle equals half the difference of the arcs: (80 - 30)/2 = 25 degrees. Had the same two arcs been intercepted by chords crossing inside, the angle would be half their sum, (80 + 30)/2 = 55 degrees. The vertex position, not the arcs, decides sum versus difference.
Common Traps
- Sum vs. difference: inside the circle uses the sum of arcs; outside uses the difference. Fixing the vertex first prevents this slip.
- Radius vs. diameter in chord problems: the hypotenuse of the chord right triangle is the radius, not the diameter.
- Tangent slope: the tangent is perpendicular to the radius, so its slope is the negative reciprocal of the radius slope, not the same slope.
From external point P, tangent segments PA = 2x + 5 and PB = 3x - 1 are drawn to a circle. What is the value of x?
Two chords intersect inside a circle. One chord is divided into segments of length 4 and 6; the other into segments of length 3 and x. What is x?
A tangent and a chord meet at a point on a circle, forming a 40-degree angle. What is the measure of the intercepted arc?