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Circle with center O.  AB is a diameter, CO is perpendicular to AB.  PO = 25 and PQ = 8.  Find the radius.

 

 Dec 8, 2020
 #1
avatar+114221 
+1

By  the intersecting  chord theorem

 

PQ * CP  = PB * AP

 

PB  = r + 25

AP  = r -25

 

So

 

8 * CP = (r + 25) (r -25)

 

CP  =  (r^2 -625)  / 8      →  CP^2  =  (r^2 -625)^2  / 64

 

Therefore

 

CP^2 - PO^2  =  CO^2  = r^2 

 

(r^2 - 625)^2 /64  -  25^2  =  r^2

 

r^4 - 1250r^2  + 390625  -  40000  =  64r^2

 

r^4  - 1314r^2  + 350625  =  0

 

 

Solving  this  produces  that  the possible positive values for  r   ≈  19.296   or  r ≈ 30.686

 

r  must  be > 25   ....so....r ≈  30.686

 

cool cool cool

 Dec 8, 2020
 #3
avatar+846 
0

Good job, Phill!!!smiley

jugoslav  Dec 10, 2020
 #2
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8 * [sqrt(r2 + 252)] = (r - 25)(r + 25) smiley

 Dec 9, 2020

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