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# Challenge problems: radius & tangent

Solve two challenging problems that apply properties of tangents to find the radius of a circle with a tangent.

## Problem 1

A, C, with, \overleftrightarrow, on top is tangent to circle O at point C.
A circle centered around point O. Segment O C is a radius of the circle. Point A lies outside the circle, and line A C is a line that could potentially be tangent to circle O. A line segment connects point A to point O and intersects the circle at point B. Line segment A O, line segment O C, and line A C create the triangle A O C. Side A O is broken into two line segments, A B and B O. Segment A B is two units. Segment A C is four units.
What is the length of start overline, O, C, end overline?
units

## Problem 2

A, C, with, \overleftrightarrow, on top is tangent to circle O at point C.
A circle centered around point O. Segment O C is a radius of the circle. Point A lies outside the circle, and line A C is a line that could potentially be tangent to circle O. A line segment connects point A to point O and intersects the circle at point B. Line segment A O, line segment O C, and line A C create the triangle A O C. Side A O is broken into two line segments, A B and B O. Segment A B is eighteen units. Segment A C is thirty units.
What is the length of start overline, O, C, end overline?
units

## Want to join the conversation?

• Hey guys, do you guys have any tricks that I could use on a test when doing these types of problems? Any way to do it quicker without all the calculations?
• Depending on the test (ACT for example) you will be able to use a calculator. On problem 2, I skipped a few steps by leaving the square integers as they were; as in I wrote 30^2 instead of bothering to calculate that it equals 900. Proceeded with my algebra, then came up with (OC)= ((30^2)-(18^2))/36, which I could just type into my calculator to get 16. As you practice more and more you'll get little nuances like that which work for you to speed up your game, and it all comes from having fun with the numbers and the math and practice.
• why is there a 36 i didn't get a 36 when i did my answer
(1 vote)
• We're just squaring binomials (x+a)^2 = x^2+2ax+a^2
• how do you get 36OC from the second problem. wouldn't you get 324 again because you are putting 18 to the power of 2
• Are they sure it's actually solvable?
Because to me it looks like (talking about the first problem) they're pulling that 4OC out of nowhere, along with adding together OC and 2 in their squared forms, which would not yield the same result as adding them together and THEN squaring them.

Okay, was a bit frustrated writing that, but all that said, could anyone please better explain where some of those extra numbers come from, and how everything ends up working like a regular Pythagorean problem?
(1 vote)
• We're finding the square of two things added together, so we're using the fact that (a+b)^ 2 = a^2 + b^2 + 2ab
so (OC + 2)^2 = (OC)^2 + (2)^2 + 2(OC)(2)
= OC^2 + 4OC + 4
yay!
• i dont get this
• You can use the Pythagorean theorem:
Set OC and OB as x, they are equal because all radii are congruent.
OC is perpendicular to AC (line tangent to a circle is perpendicular to the radius drawn to the point of tangency), making OAC a right triangle.
OA is the hypotenuse, OC and AC are the legs. So you can set the equations for the Pythagorean Theorem:
1) 4^2+x^2=(2+x)^2
16+x^2=4+4x+x^2
16=4+4x
12=4x
x=3
2) 30^2+x^2=(18+x)^2
900+x^2=324+36x+x^2
900=324+36x
576=36x
x=16
(1 vote)
• Does anyone have a simple explanation for this PLEASE