one circle has radius 5 and its centre (0,5). A second circle has radius of 12 and its centre (12,0). What is the length of the radius of a third circle which passes through the centre of the second circle and both the points of intersection of the first two circles.

For first circle centre  is  (0 , 5) and radius is 5 so equation of the circle,
(x - 0)+ (y - 5)2 = 52 
x+ y- 10y = 0
For second circle centre  is (12, 0) and radius is 12 so,
(x - 12)+ (y - 0)2 =  122 
x+ y2  - 24x = 0
Now third circle is passing through these two circles intersection points so it will be 
 x+ y- 10y + λ(x+ y - 24x) = 0
Also  third circle is passing through point (12,0) so putting this point  in above equationwe get  λ = 1 and put this value we get  equation of  third circle,
x+ y- 12x + 5= 0.
This gives centre (6 , 5/2) and radius = 6.5

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