AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

The equation of the two circles are,
x2+y2+2x+4y−20=0
x2+y2+6x−8y+10=0
This implies,
2g1g2+2f1f2=6−16
=−10
=c1+c2
The circle intersects orthogonally so have a common tangent.
Now the equation of common chord is,
2x−6y+15=0
So, the length of common chord is,
2√25−

(−2+12+15)2

4+36

=2√25−

625

40

=2√

1000−625

40

=

5√3

√2