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The following circle (fig.) has two secants from a common point C, namely AC and BC. The circumference of the circle intersects these two secants at E and D respectively. If AE = 6 , BD = 5 and DC = 3, then what is the length of seg DC ?

 

 Jan 8, 2021
 #1
avatar+114361 
+2

I think you want  CE , not DC    (DC  = 3)

 

We  have  the secant-secant theorem

 

CE (CA)    =  CD (CB)

 

x ( 6 + x)  =  3 ( 3 + 5)

 

6x + x^2  =  3 * 8

 

x^2 + 6x  =  24    complete the square on x

 

Take (1/2) of 6 = 3  

Square it =  9   and add to both sides

 

x^2 + 6x  + 9 =  24 +  9

 

(x + 3)^2  =  33      take the  positive  root

 

x +  3   =sqrt (33)

 

x =  CE  =   sqrt (33)  -  3   ≈  2.7446

 

 

cool cool cool

 Jan 8, 2021
 #2
avatar+188 
+2

We can use the POP formula. 3(3+5)=x(x+6) so 24=x^2+6x meaning x^2+6x-24=0. We can use the quadratic formula to get x=-6+-(sqrt132)/2 which means x=sqrt33-3 because the length can't be negative.

:D

 Jan 8, 2021

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