AP EAMCET Engineering 2018 Apr 23 Shift 2 Paper

The distance between parallel lines 3x−4y−10=0 and3x−4y+30=0 is the length of diameter of required circle so radius is,

1

2

40

√9+16

=4
And midpoint of the intersection of the line 3x−4y−10=0 and3x−4y+30=0,x+2y=0 is the center of required circle to center is,
(

2−6

2

,

−1+3

2

)=(−2,1)
Then the equation of required circle is,
x2+y2+4x−2y−11=0