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Q.

The differential equation of the family of circles passing through the fixed points (a, 0) and (- a, 0), is 

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a

y1y2x2+2xy+a2=0

b

none of these

c

y1y2+xy+a2x2=0

d

y1y2x2+a2+2xy=0

answer is C.

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Detailed Solution

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Let the equation of the family of circles be

x2+y2+2gx+2fy+c=0

It passes through (a, 0) and (- a, 0). Therefore, 

a2+2ag+c=0 and a22ag+c=0

Solving these two equations, we get c=a2 and g=0

Substituting the values of c and gin (i), we get

x2+y2+2fya2=0

It is a one parameter family of circles. 

Differentiating with respect to x, we get 

2x+2yy1+2fy1=0f=x+yy1y1

Substituting the value off in (ii), we get

y1y2x2+a2+2xy=0

This is the required differential equation

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