Tangent lines

[bnefriends]

My geometry workbook is telling me "in each diagram, AB (with --> above AB) is tangent to (circle with a dot in the center) C at B. Find the value of x." For problem number 4, point C is the midpoint of a circle, and triangle ABC overlaps the circle. Side AB, the longer leg, is 8 cm. BC, the shorter leg, is X cm. AC, the hypotenuse, is 2 cm outside of the circle and X cm inside the circle. If anybody has Pearson Hall Geometry Daily Skills and Practice workbook, I'm having trouble with the problems on 11-1, page 131-132. I've tried using the textbook, but it's not helping me with some of these problems.

 

[Mrspi]

My geometry workbook is telling me "in each diagram, AB (with --> above AB) is tangent to (circle with a dot in the center) C at B. Find the value of x." For problem number 4, point C is the midpoint of a circle, and triangle ABC overlaps the circle. Side AB, the longer leg, is 8 cm. BC, the shorter leg, is X cm. AC, the hypotenuse, is 2 cm outside of the circle and X cm inside the circle. If anybody has Pearson Hall Geometry Daily Skills and Practice workbook, I'm having trouble with the problems on 11-1, page 131-132. I've tried using the textbook, but it's not helping me with some of these problems.

You're expected to use the theorem which says "If a line is tangent to a circle, then it is perpendicular to a radius of the circle drawn to the point of tangency."

Ray AB is tangent to circle C at point B, so B is the "point of tangency." Segment CB is a radius of circle C drawn to the point of tangency. So, segment CB is perpendicular to ray AB at B....which means that angle ABC is a right angle, and triangle ABC is a right triangle. AB and BC are the two legs of that right triangle, and AC is the hypotenuse. You can use the Pythagorean Theorem, which says that in ANY right triangle,

leg[sup:lu106p5v]2[/sup:lu106p5v] + leg[sup:lu106p5v]2[/sup:lu106p5v] = hypotenuse[sup:lu106p5v]2[/sup:lu106p5v]

In your right triangle,
(AB)[sup:lu106p5v]2[/sup:lu106p5v] + (BC)[sup:lu106p5v]2[/sup:lu106p5v] = (AC)[sup:lu106p5v]2[/sup:lu106p5v]

You are told that AB = 8, BC = x, and AC = x + 2.

Substitute those expressions into the equation, and solve for x.

If you're still having trouble, please repost and show us ALL of the work you've done so we can determine exactly where it is that you need further help.