Answer:
TL = 8 in.
CX = 3 in.
Explanation:
TK is a tangent to circle O at point K. That makes K a right angle.
Triangle TKO is a right angle.
All units of length are inches. I will leave out in. until the end for simplicity.
TO = 10
KO = 6 in.
Using the Pythagorean theorem,
(TK)² + (KO)² = (TO)²
(TK)² + 6² = 10²
TK = √(100 - 36)
TK = 8
We know all side lengths of triangle TKO.
Now look at triangle CXT.
CX is tangent to the circle. That makes <CXT a right angle.
Triangle CXT is a right triangle.
<KTO of triangle TKO and <CTX of triangle CXT are congruent because they are the same angle.
Since the triangles are right triangle, they also have another pair of congruent angles, the two right angles, TKO and CXT. By AA similarity, triangles TKO and TXC are similar triangles.
TO = TX + OX
OX is a radius of circle O. OX = 6.
10 = TX + 6
TX = 4 in.
Now we use a proportion of side lengths of the similar triangles.
Triangle TLO is congruent to triangle TKO since TL = TK (tangents), KO = LO (radii) and both are right triangles.
TL = TK = 8 in.
TX/TK = CX/KO = TC/TO
4/8 = CX/6
1/2 = CX/6
CX = 3 in.