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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)
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
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