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# need help on this

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In circle O as shown above, AB=2 cm and DC=6 cm, ∠BAC=∠ACD=45∘. What is the radius of circle O? Jun 20, 2020

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Let V be the midpoint of arc(AB) and Z the midpoint of arc(CD).

Draw VWXYZ, where W is the point of intersection with AB (W will be the midpoint of AB);

X is the point of intersection with AC;

Y is the point of intersection with CD (Y will be the midpoint of CD).

VWXYZ will be perpendicular to both AB and CD.

Since AB = 2, AW = 1 and WX = 1.

Since CD = 6, CY = 3 and XY = 3.

Call XW = a and call YZ = b.

Since WXYZ is a diameter and AB is a chord of the circle:  AW · WB  =  VW · WZ.

--->     1 · 1  =  a · (1 + 3 + b)     --->     1  =  a · (4 + b)     --->     4a + ab  =  1     --->     ab  =  1 - 4a

Since WXYZ is a diameter and CD is a chord of the circle:  CY · YD  =  YZ · VY.

--->     3 · 3  =  b · (1 + 3 + a)     --->     9  =  b · (4 + a)     --->     4b + ab  =  9     --->     ab  =  9 - 4b

Combining these two results:  1 - 4a  =  9 - 4b    --->     4b - 4a  =  8     --->     b - a  =  2

--->     b  =  2 + a

Since     a · (4 + b)  =  1     --->     a · (4 + (2 + a) )  =  1     --->     a · (6 + a)   =  1

--->     6a + a2  =  1     --->     a2 + 6a - 1  =  0

--->     a  =  -3 + sqrt(10)      [quadratic formula]

Since      b · (4 + a)  =  9     --->     b · (4 + (b - 2) )  =  9     --->     b · (2 + b)   =  9

--->     2b + b2  =  9     --->     b2 + 2b - 9  =  0

--->     b  =  -1 + sqrt(10)     [quadratic formula]

The diameter will be:     (a) + (1) + (3) + (b)  =  ( -3 + sqrt(10) ) + (1) + (3) + ( -1 + sqrt(10) )

=  2sqrt(10)