McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
8. Congruent and Similar Solids
Continue to next subchapter

Exercise 41 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.

5

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* (x + 5+x)=5* (5+ 5+x) Let's solve this equation for x.
x * (x+ 5+x) = 5 * (5+ 5+ x)
â–Ľ
Solve for x
x * (2x + 5) = 5 * (10 + x)
2x^2 + 5x = 5 * (10 + x)
2x^2 + 5x = 50 + 5x
2x^2 = 50
x^2 = 25
x = sqrt(25)
x = 5
Note that, when solving the above equation, we only kept the principal root because x represents a length and therefore is a positive number.