Help with a geometry question..?

Question

2. Given:

• circle O

• Tangent line PT with T the intersection point

• Secant line PR with intersection points J and R where J is between P and R

• Point N is between J and R

• JR is not a diameter

• PT = 12, PJ = 8, ON = 4, NJ = 6

Find the radius.

Please explain how you got it.. Thanks!

Answer 1

Draw a diagram.

By the power chord theorem:

(PJ)(PR) = (PT)²

(PR) = (PT)² / (PJ) = 12² / 8 = 144/8 = 18

Solve for chord JR.

JR = PR - PJ = 18 - 8 = 10

Let

M = midpoint between J and R.

JM = JR / 2 = 10/2 = 5

On chord JR we have:

MN = JN - JM = 6 - 5 = 1

Now we have a right triangle OMN, with ON the hypotenuse.

(OM)² = (ON)² - (MN)² = 4² - 1² = 16 - 1 = 15

Now we have a second right triangle OMJ with OJ the hypotenuse and also the radius of circle O.

(OJ)² = (OM)² + (JM)² = 15 + 5² = 15 + 25 = 40

OJ = √40 = 2√10

The radius is 2√10.