AP EAMCET Engineering 26 Apr 2017 Shift 1 Paper

Consider the given circles,
S1≡x2+y2+2x+2y+1=0
And,
S2≡x2+y2+4x+6y+4=0
The equation of common chord of these two circles,
s1−s2=0
−2x−4y−3=0
2x+4y+3=0
The equation of the required circle is,
S1+λ(s1−s2)=0
(x2+y2+2x+2y+1) +λ(2x+4y+3)=0
x2+y2+2x(1+λ)+2y(2λ+1) +3λ+1=0
Now, 2x+4y+3=0 is the diameter of the circle.
So,
2(−(1+λ))+4(−(2λ+1))+3=0
−2−2λ−8λ−4+3=0
10λ=−3
λ=−

3

10

Therefore, the equation of the required circle is,
x2+y2+2x(1−

3

10

)+2y(

−6

10

+1) −

9

10

+1=0
x2+y2+

14x

10

+

8y

10

+

1

10

=0
10x2+10y2+14x+8y+1=0