The correct option is B equation of the circle which is orthogonal to given circles isx2+y2−6x−4y−14=0
Given circles are
S1:x2+y2+3x+2y+1=0
S2:x2+y2−x+6y+5=0
S3:x2+y2+5x−8y+15=0
Equations of two radical axes are
S1−S2=0⇒4x−4y−4=0⇒x−y−1=0
And
S2−S3=0⇒−6x+14y−10=0⇒3x−7y+5=0
Solving them, we get the radical centre as (3,2)
Also, if r is the length of the tangent drawn from the radical centre (3,2) to any one of the given circles, say S1, we have
r=√S1=√32+22+3.3+2.2+1⇒r=√27
Hence, (3,2) is the centre and √27 is the radius of the circle intersecting them orthogonally.
∴ Its equation is
(x−3)2+(y−2)2=27⇒x2+y2−6x−4y−14=0