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.

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

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

Lets say OC = a (a + 18)^2 (a + 18) * (a + 18) a^2 + 36a +324 http://www.khanacademy.org/math/algebra/introduction-to-polynomial-expressions/multiplying-binomials-2/v/multiplying-binomials

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?

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!

why is there a 36 i didn't get a 36 when i did my answer

We're just squaring binomials (x+a)^2 = x^2+2ax+a^2

How would I be able to solve problem 1 without knowledge of the radius of circle O?

Math 101: If you don't know something, let it be something! Let radius of circle O be r. By Pythagorean Theorem, 4² + r² = (2 + r)² 16 + r² = 4 + 4r + r² 4r = 12 r = 3 Thus OC is 3 units. Also the problem asks for OC which is the radius, of course you don't have the radius.

Main content

Course: Integrated math 2 > Unit 11

Lesson 10: Properties of tangents

Challenge problems: radius & tangent

Google Classroom

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

Problem 1

\overset{\leftrightarrow}{A C}

is tangent to circle

O

at point

C

What is the length of $\overset{―}{O C}$ ‍?

units

Because

\overset{\leftrightarrow}{A C}

is tangent to the circle at point

C

, the radius going to point

C

is perpendicular to the tangent line. This means that the measure of

∠ A C O

90^{\circ}

We can use the Pythagorean theorem to solve for the length of

\overset{―}{O C}

We will need to know the length of

\overset{―}{O A}

We know that

A B = 2

B O = O C

because it is a radius of the circle.

$O A = A B + B O$ ‍
$O A = 2 + O C$ ‍

Now that we have an expression for the length of

A O

, we can solve for

O C

$\begin{aligned} {A C}^{2} + {O C}^{2} & = O A^{2} \\ 4^{2} + {O C}^{2} & = (O C + 2)^{2} \\ 16 + {O C}^{2} & = {O C}^{2} + 4 O C + 4 \\ 12 & = 4 O C \\ 3 & = O C \end{aligned}$ ‍

The length of $\overset{―}{O C}$ ‍ is $3$ ‍ units.

Problem 2

\overset{\leftrightarrow}{A C}