AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

The point P is taken as.
S1(a,b)=S2(a,b)
a2+b2+2ga+2fb+c =a2+b2+

3

2

a+4b+c=0
a(2g−

3

2

)+b(2f−4)=0
This is locus of radical axis.
So. x(2g−

3

2

)+y(2f−4)=0 is the radical axis of the given circles.
This touched the curve of equation x2+y2+2x+2y+1=0
The radius of the circle is given as,
r=√12+12−1=1
The centre of the circle is (-1,-1)
The radius of thee circle is equal to distance between centre and touching point.

|( 3 2 −2g)+(2f−4)|

√( 3 2 −2g)2+(2f−4)2

=1
Squaring both sides we get,
2(

3

2

−2g)(2f−4)=0

3

2

−2g=0 or 2f−4=0
g=

3

4

or f=2