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# Help with circles

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A and B are two points on a circle with center O, and C lies outside the circle, on ray AB. Given AB = 24, BC = 30, and OA = 16, find OC.

Jun 4, 2021

#1
+122
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OC=$$\sqrt{1876}$$

Proof

The proof is difficult to explain concisely and without a diagram, but basically, draw AB, O, and the perpendicular from O to the midpoint of AB. The intersection will be labeled D.  Thus, AD, DO, and OA will form a right triangle, hypotenuse of 16 and one leg of AD of 12.  The other leg (DO) is $$\sqrt{112}$$ through the Pythagorean theorem.  Now, consider the right triangle DCO.  DC= 30+12 = 42, and DO =$$\sqrt{112}$$.  From the Pythagorean theorem, OC = $$\sqrt{112+42^2}$$ or $$\sqrt{1876}$$.

Have a great day!

Jun 4, 2021
#3
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"The proof is difficult to explain concisely and without a diagram."

This reminds me of the quote "I have discovered a truly remarkable proof of this theorem which this margin is too small to contain"

#2
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Here is  what EnchantedLava is saying

AB is a chord on the  circle.....call the mid point D.....then AD =   DB =   24/2 = 12

And  we can form  right triangle   AOC with   AO =   16  = the hypotenuse,  and leg AD =  12

And OD  will  form the other leg   such that    AO^2 - AD^2 =  OD^2

16^2  - 12^2   =     112  =  OD^2

And we can form another right triangle  DOC such that OD and DC are legs.....and  DC  = ( DB + BC)  = 12 + 30 = 42

And OC will form the hypotenuse of this triangle....such that

OC  =  sqrt  ( OD^2  + DC^2)  = sqrt (112  + 42^2)  = sqrt (1876)

See the image here :

Jun 4, 2021