Mastering Circle Tangent: Understanding Lines in a Circle and Tangent Lines to a Circle

From the graph, it can be seen that $\overline{A D}$ and $\overline{A E}$ are tangent segments with a common endpoint outside $⊙ P .$ By the External Tangent Congruence Theorem, $\overline{A E}$ and $\overline{A D}$ are congruent.

Smilarly,

\overline{B E}

and

\overline{B F},

and

\overline{C F}

and

\overline{C D}

are congruent tangent segments.

\overline{A D} ≅ \overline{A E} \overline{B E} ≅ \overline{B F} \overline{C F} ≅ \overline{C D} \Leftrightarrow A D = A E B E = B F C F = C D

Using the Segment Addition Postulate and the given lengths, a system of equations with three equations and three unknowns can be written.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ A E + B E = A B B F + C F = B C C D + A D = C A \Rightarrow ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ A E + B E = 12 B E + C F = 10 C F + A E = 6 (I) (II) (III)

To find

A E,

the Elimination Method can be used. Start by multiplying the second equation by

- 1 .

- 1 (B E + C F) = - 1 (10) ⇕ - B E - C F = - 10

Adding this equation to the first equation will eliminate

B E .

A E + B E + (- B E - C F) = 12 + (- 10)

Simplify

AddNeg

$a + (- b) = a - b$

A E + B E - B E - C F = 12 - 10

SubTerm

Subtract term

A E - C F = 2

Since

- C F

and

C F

are additive inverses, adding this equation to the third equation will eliminate

C F,

and thus give

A E .

C F + A E + (A E - C F) = 6 + (2)

Simplify

RemovePar

Remove parentheses

C F + A E + A E - C F = 6 + 2

AddSubFrac

Add and subtract fractions

2 A E = 8

DivEqn

$LHS / 2 = RHS / 2$

A E = 4

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Challenge

Investigating Properties of Tangents to a Circle

Discussion

Drawing a Tangent Line to a Circle Through an Outer Point

Discussion

External Tangent Congruence Theorem

Example

External Tangent Congruence Theorem

Hint

Solution

Closure

Tangents to Circles in Real Life

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