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9. Circles
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United States of AmericaFL
Grade 912

9.06 Lengths in chords, secants, and tangents

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Practice
Lesson

Concept summary

The following two theorems relate to the lengths of segments formed by secants and tangets to a circle from a common external point:

Secant secant theorem

If two secant segments share the same endpoint outside a circle, then the product of the length of one secant segment and the length of its external segment is equal to the product of the length of the other secant segment and the length of its external segment.

A circle with two segments that are secants to the circle, and intersecting at a point S outside the circle. One of secants is divided into segments of length m, and length n. The segment of length m is a chord of the circle, and the segment with length n is an external segment. The other secant is divided into segments of length a, and length b. The segment of length a is a chord of the circle, and the segment of length b is an external segment.

For the diagram shown, the secant secant theorem says that\left(n + m\right)n = \left(b+a\right)b

Tangent secant theorem

If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the length of the secant segment and the length of its external segment is equal to the square of the length of the tangent segment.

A circle with a segment that is secant to the circle, and another segment that is tangent to the circle. The secant and the tangent intersect at a point T outside the circle. The tangent segment has a length of x, and has endpoints at T, and at the point of tangency. The secant has an endpoints at T, and at a point on the circle. It intersects the circle at two points. The secant is divided into two segments of length y, and length z. The segment with length y is a chord, and the segment with length z is an external segment.

For the diagram shown, the tangent secant theorem says thatx^2=\left(z+y\right)z

Intersecting chord theorem

If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

A circle with points A, G, B, and F placed clockwise on the circle. Chords A B and F G are drawn, and intersect at a point P in the circle.

For the diagram shown, the intersecting chord theorem says that:

{AP} \cdot {PB} = {FP}\cdot {PG}

Worked examples

Example 1

Given AB = 5, CB = 3 and CD = 2x+5. Solve for x.

A circle and with points D, C, and A placed clockwise on the circle. Secant ray B D passing through C is drawn. C is between B and D. A ray from B to A is drawn. B A is tangent to the circle at A.

Approach

We have been given lengths of all the segments needed for the tangent secant theorem, so we can use that to construct an equation and solve for x.

Solution

\displaystyle AB^2\displaystyle =\displaystyle \left(BC + CD \right) BCFormula
\displaystyle 5^2\displaystyle =\displaystyle \left(3+(2x+5)\right)\cdot \left(3\right)Substitution
\displaystyle 25\displaystyle =\displaystyle \left(2x+8\right)3Combine like terms
\displaystyle 25\displaystyle =\displaystyle 6x+24Distribute the 3
\displaystyle 1\displaystyle =\displaystyle 6xSubtract 24 from both sides
\displaystyle \dfrac{1}{6}\displaystyle =\displaystyle xDivide both sides by 6
\displaystyle x\displaystyle =\displaystyle \dfrac{1}{6}Symmetric property of equality

Therefore, x=\dfrac{1}{6}.

Example 2

Given DE = 20, CD = 11 and BC = 13. Find the length of the segment AB.

A circle with points D, E, A, and B are on the circle. Secant segment C E which passes through D, and secant segment C A which passes through B are drawn.

Approach

We have been given lengths of all the segments needed for the secant secant theorem, so we can use that to construct an equation and solve for x.

Solution

\displaystyle \left(DE + CD\right)CD\displaystyle =\displaystyle \left(BC + AB\right)BCFormula
\displaystyle \left(20 + 11\right)11 \displaystyle =\displaystyle \left(13+AB\right)13Substitution
\displaystyle 341\displaystyle =\displaystyle \left(13+AB\right)13Combine like terms and simplify
\displaystyle 341\displaystyle =\displaystyle 13^2 + 13 \cdot ABDistribute the 13
\displaystyle 341\displaystyle =\displaystyle 169 + 13 \cdot ABSimplify
\displaystyle 172\displaystyle =\displaystyle 13 \cdot ABSubtract 169 from both sides
\displaystyle \dfrac{172}{13}\displaystyle =\displaystyle ABDivide both sides by 13
\displaystyle AB\displaystyle =\displaystyle \dfrac{172}{13}Symmetric property of equality

Therefore, AB=\dfrac{172}{13}.

Example 3

Determine the length of \overline{AC}, if AB = 7\, \text{cm}, AD = AE, and AE = 3\, \text{cm}.

A circle with points C, E, D, and B placed clockwise on the circle. Chords C D and B E are drawn, and intersect at a point A in the circle.

Approach

We can use the intersecting chord theorem. We also know that AD = 3\, \text{cm} since AD = AE.

Solution

\displaystyle AB \cdot AE \displaystyle =\displaystyle AD \cdot ACIntersecting chord theorem
\displaystyle 7 \cdot 3\displaystyle =\displaystyle 3 \cdot ACSubstitute known values
\displaystyle 7\displaystyle =\displaystyle ACSimplify

The length of AC is 7\, \text{cm}.

Reflection

Since we know that AE = AD, then the intersecting chord theorem tells us that AC = AB as well. In this case, both AC and AB measure 7\, \text{cm}.

Outcomes

MA.912.GR.6.1

Solve mathematical and real-world problems involving the length of a secant, tangent, segment or chord in a given circle.

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