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Question

A system of circles is said to be coaxial when every pair of circles has the same radical axis. For coaxial circles, we note that
(1)The centers of all coaxial circles lie in a straight line, which is to the common radical axis
(2)Circles passing through two fixed points form a coaxial system with a line joining the points as a common radical axis.
(3)The equation to a coaxial system whose two members are S1=0 & S2=0 is given by S1+λS2=0,λ is parameter.
If we take line of centres as x -axis & common radical axis as y-axis, then the simplest form of equation of coaxial circles is given by x2+y2+2gx+c=0 where g is variable & c is constant
If g=±c then radious g2c vanishes & the circle become a point circle. The points (±c,0) are called the limiting points of the system of coaxial circle given by x2+y2+2gx+c=0
On the basis of above information answer the following question:
Equation of circle through the origin & belonging to the coaxial system, of which the limit points are (1,2) & (3,4) is

A
x2+y2+4x+3y=0
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B
x2+y24x+3y=0
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C
x2+y2+4x3y=0
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D
x2+y24x3y=0
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Solution

The correct option is C x2+y2+4x3y=0
As limiting points are (1,2) & (3,4)

Point circles of the coaxial system are given by

(x+1)2+(y2)2=0 and

(x3)2+(y4)2=0

or x2+y2+2x4y+5=0 & x2+y2+6x8y+25=0

Equation of coaxial system is given by

(x2+y2+2x4y+5)+λ(x2+y26x8y+25)=0 .....(1)

Now equation (1) passess through (0,0)

So ,λ=15

From equation (1) we have

5(x2+y2+2x4y+5)(x2+y26x8y+25)=0

5(x2+y2+2x4y+5)(x2+y26x8y+25)=0

x2+y2+4x3y=0

Hence choice (C) is correct answer.

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