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Question

The equation of a circle passing through the point (1,1) and the point of intersection of the circles
x2+y2+13x−3y=0 and 2x2+2y2+4x−7y−25=0 is

A
4x2+4y2+30x13y25=0
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B
4x2+4y2+30x13y+25=0
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C
4x24y230x+13y25=0
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D
4x24y2+30x13y25=0
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Solution

The correct option is D 4x2+4y2+30x13y25=0
the equation of any curve through the points of intersection of the circles (1) and (2) will be
(x2+y2+13x3y)+k(2x2+2y2+4x7y25)=0
It is given that equation of circles passes through the point (1,1)
So,
x=1 and y=1
Substitute these value in an above equation, we get
(1+1+133)+k(2+2+4725)=0
1224k=0
k=12
Now, Subsitute the value of k in an first equation, we get
(x2+y2+13x3y)+12(2x2+2y2+4x7y25)=0
2x2+2y2+15x13y2252=0
4x2+4y2+30x13y25=0

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