Circle

Asked by thebluebloo | 16th Dec, 2008, 03:10: PM

Expert Answer:

let the center is (h , k) which lie on line x+y-9=0 so h + k = 9...(1)

as a  line perpendicular to line 2x-y+1=0 and passing through (2 , 5) is normal to the circle and passes through the center so eqn of the normal is x + 2y = 12 .    point (h,k) also lie on this line so h + 2k = 12.........(2)                     

solving 1 and 2 we get (6,3) as center of the circle  and length of perpendicular from (6,3) on line 2x-y+1=0 is the radius of the circle which is (12-3+1)/(22 + 12) = 55

som eqn of circle is (x-6) + (y-3)2= 50

Answered by  | 16th Dec, 2008, 07:53: PM

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