McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
8. Congruent and Similar Solids
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Exercise 42 Page 870

Whenever there are two points of intersection along a line through a point and a circle, the product of the lengths of the segments created by the point outside the circle and the points of intersection is constant for any line.

4

Practice makes perfect

Often there are two points of intersection along a line through a point and a circle. These points of intersection create two different segments between the point outside the circle and the circle — to the first point of intersection and to the second point of intersection.

The product of the lengths of the segments created by the point outside the circle and the points of intersection is constant for any line. Therefore, the products of the secants and their outer segments are equal. x* (5+x)=3* (9+3)Let's solve this equation for x.
x * (5+x) = 3 * (9+ 3)
x * (5+x) = 3*(12)
x * (5+x) = 36
5x + x^2 = 36
5x + x^2 - 36 = 0
x^2 + 5x - 36 = 0
We obtained a quadratic equation. Let's identify the values of a, b, and c. x^2 + 5x - 36 = 0 ⇕ 1x^2+ 5x+( - 36)=0 We see that a = 1, b = 5, and c = - 36. Next, we will substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
â–Ľ
Solve using the quadratic formula
x=- ( 5)±sqrt(5^2-4( 1)( - 36))/2( 1)
x=-5±sqrt(25 - 4(1)(-36))/2(1)
x=-5±sqrt(25 + 144)/2
x=-5±sqrt(169)/2
x = -5 ± 13/2
We will find the values for x by using the positive and negative signs.
x=-5 ± 13/2
x=-5 + 13/2 x=-5 - 13/2
x=8/2 x=- 18/2
x=4 x=- 9

Since x is the value of a length, x has to be a positive number. Therefore, x=4.