Gautam H.

asked • 01/29/16

find the locus

if the two circles x2 + y2 + 2gx +2fy + c = 0 and x2 + y2 - 2fy - c = 0 have equal radius then locus of (g, f) is
 
(I)x2 + y2 = c2
(ii)x2 - y2 = c2
(iii)x - y2 = c2
(iv)x2 +y2 = 2c2

Gautam H.

how (g,f) = (x,y)
 
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01/31/16

Michael P.

Gautam,
 
It would have been simpler for you if the author of this problem had written the answers in terms of g and f, for instance, (i) g2 + f2 = c2, but the equations, such as g2 = 2c or g2 + f2 = 2c, produced by this type of problem are plotted in the (x,y) plane by replacing g with x and f with y, that is, (g,f) = (x,y) or (x,y) = (g,f). So, you have to make those substitutions to plot the answer, and more importantly for this problem, to recognize the answers (as functions in the same (x,y) plane as the original equations).
 
BTW, Gautam, check the signs on the 2fy terms and that the constant term, c, is not c2. If 2fy in one of the original equations had the opposite sign and the constant term in both equations was c2, then (iv) would be the correct answer. None of the other answers would work because the signs on the constant terms result in a factor of 2c (or 2c2) in the answer.
 
Michael.
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01/31/16

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Michael P. answered • 01/29/16

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Gautam,
 
The canonical form of a circle of radius r is (x - h)2 + (y - k)2 = r2.
Use completing the square for x and y in the first equation and for y in the second equation to get both equations into canonical form so that you can equate the radiuses (radii) and produce an equation for g and f.
 
x2 + y2 + 2gx +2fy + c = 0
 
x2 + 2gx +y2 + 2fy + c = 0
 
x2 + 2gx + g2 + y2 + 2fy + f2 + c - g2 - f2 = 0
 
(x + g)2 + (y + f)2 + c - g2 - f2 = 0
 
(x + g)2 + (y + f)2 = g2 + f2 - c = r2
 
and
 
x2 + y2 - 2fy - c = 0
 
x2 + y2 - 2fy + f2 - c - f2 = 0
 
x2 + (y - f)2 - c - f2 = 0
 
x2 + (y - f)2 = c + f2 = r2
 
Equating the radius squared in both equations:
 
r2 = g2 + f2 - c = c + f2 
 
g2 = 2c
 
The locus of (g,f) is the parabola g2 = 2c or x2 = 2c given (x,y) = (g,f).
 
The answer is "None of the above"!
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