Answer:
Explanation:
You want the segment lengths DA and DE of the hypotenuse in the triangle shown in the figure.
Right triangle
The radius to a point of tangency always makes a right angle with the tangent. This is a right triangle with legs 8 and 15, so you know from your knowledge of Pythagorean triples that the hypotenuse is 17.
DA = 17
DE = 17 -8 = 9
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Additional comment
In case you haven't memorized a few of the useful Pythagorean triples, {3, 4, 5}, {5, 12, 13}, {7, 24, 25}, {8, 15, 17}, you can always figure the missing side length of a right triangle using the Pythagorean theorem.
It tells you the sum of the squares of the legs is the square of the hypotenuse:
AC² +CD² = DA²
8² +15² = DA²
64 +225 = 289 = DA²
DA = √289 = 17
Of course, AE is the radius of the circe, 8, so ...
AE + DE = DA
8 +DE = 17
DE = 17 -8 = 9
Alternatively, you can solve this using the relation between tangents and secants. If the line DA is extended across the circle to intersect it again at X, then ...
DC² = DE·DX
15² = DE·DX = DE(DE +16) . . . . . . . EX is the diameter, twice the radius of 8
DE² +16DE -225 = 0
(DE +25)(DE -9) = 0 . . . . factor
DE = 9 . . . . the positive solution
DA = 9 +8 = 17
We like the Pythagorean theorem solution better, as the factors of the quadratic may not be obvious.