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Question

Find the equations of a circle which passes though the point ( 2 , 0 ) and whose center is the limit of the point of intersection of the lines 3x + 5y = 1 , (2 + c ) x + 5c2 y = 1 tends to 1

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Solution

Eliminating y, we get
(3c22 c - 2)x = c2 - 1 x=(c1)(c+1)(c1)(3c+2)=c+13c+2
limit when c 1 gives x=25 and hence y=125.
Centre ( g , f) is (25,125).
Circle is
x2+y2(45)x+(225)y+k=0
It passes through (2,0) k=(125)
Hence the required circle is
25(x2+y2)20x+60y=0.

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