TS EAMCET 5 May 2018 Shift 1 Solved Paper

Consider the equation,
x2+y2−6x−4y+13 −c2=0
x2+y2−4x−6y+13 −c2=0
The centers of the circles are C1(3,2) and C2(2,3) and, r1=c and r2=c
The two circles intersect at two points.
The equation of radial axis of common chord of the circle is given as,
−2x+2y=0
x−y=0
Draw C1M⊥AB
The length of C1M is,
C1M=|

3−2

√(1)2+(−1)2

|
=

1

√2

In ΔAC1M,
AM2=(AC1)2−(C1M)2
=c2−(

1

√2

)2
=√c2−

1

2

So, AB=2AM then,
AB=2√c2−

1

2

=√4c2−2