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Question

Let any circle S passes through the point of intersection of lines 3(y1)=x1 and y1=3(x1) and having its centre on the acute angle bisector of the given lines. If the common chord of S and the circle x2+y2+4x6y+5=0 passes through a fixed point, then the fixed point is

A
(13,32)
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B
(12,32)
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C
(12,34)
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D
(32,32)
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Solution

The correct option is B (12,32)
Given lines are
3(y1)=x1x3y+31=0(1)y1=3(x1)3x+y+31=0(2)

a1a2+b1b2=33<0
So, the acute angle bisector is
x3y+312=+(3x+y+31)2y=x
Let the centre of the circle be (α,α)

Point of intersection of (1) and (2), we get
2y+2=0y=1x=1

Now, the equation of the circle is,
(xα)2+(yα)2=(1α)2+(1α)2x2+y22αx2αy+4α2=0
Given circle is
x2+y2+4x6y+5=0

So, the common chord is,
S1S2=0(4+2α)x+(2α6)y+(74α)=0(4x6y+7)+2α(x+y2)=0
This represent family of lines, so the fixed point is the point of intersection of
4x6y+7=0 and x+y2=0y=32x=12

Hence, the fixed point is
(12,32)

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