Radius = 1/2 (diameter)
= 1/2 (10)
= 5 meters
Point Q is point outside the circle. T is a point on the circle and the point of tangency of the circle and QT.
A line tangent to the circle is perpendicular to its radius. Therefore, QT is the base, radius is the leg, and center to Q is the hypotenuse.
Using Pythagorean Theorem:
(QT)² = 20² - 5²
(QT)² = 400 - 25
(QT)² = 375
[tex] \sqrt{(QT) ^{2} } = \sqrt{375} [/tex]
QT = [tex] \sqrt{375} [/tex]
QT = [tex] \sqrt{(25)(15)} [/tex]
QT = [tex]5 \sqrt{15} [/tex] meters
