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aatish agrawal Grade: Upto college level
        

find ind the equation of the circle which pass through the focus of the parabola x^2=4ay and touches it at point (6,9)

6 years ago

Answers : (2)

gOlU g3n|[0]uS
42 Points
										

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6 years ago
vikas askiitian expert
510 Points
										

x2 = 4ay    ...........1


since circle touches the parabola at (6,9) so this point will satisfy the equation of both the curves


so , from eq 1


 36 = 36a


  a=1 , so eq 1 becomes


x2 = 4y            


now differentiating above eq       


2x = y1 = m (slope of tangent)


at (6,9) , m = 3        ..........2


let (h,k) be the center then slope of line passing through center of circle & the point of contact is given by


  m1 = (k-9)/(h-6)        .........3


mm1 = -1   or


(k-9)/(h-6) . 3 = -1        


 h + 3k = 33                      ................4


now focus of parabola is (0,1) ....


eq of circle is x2+y2 +2gx+2fy+c = 0                              (g=-h , f=-k)


 so eq of circle : x2+y2-2hx-2ky+c      


this circle passes through (0,1) & (6,9) so its eq will satisfy these points therefore


-2k+1+c = 0            ...................5


-12h-18k+117+c = 0             ....................6


from 5 & 6


3h+4k=29          ............7


solving 4 & 7 we get


(h,k) = (-9,14)


c = 27


now required eq of circle is


x2 + y2 18x - 28y + 27 = 0

6 years ago
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