AP EAPCET 23-Aug-2021 Shift 1 Paper

Given circle,
x2+y2−8x−6y+23=0
Centre of this circle is (4,3).
If the equation of radical axis of circle S1 and S2 passes through (4,3), then S2 bisect the circumference of S1.
Option (a) Let S2=x2+y2−6x−4y+9=0
Then, equation of radical axis (L) is
L:S1−S2=0
L:(x2+y2−8x−6y+23) −(x2+y2−6x−4y+9)=0
L:−2x−2y+14=0⇒L:x+y−7=0
Check whether (4,3) satisfy equation L.
LHS (4)+(3)−7=0= RHS
∴x2+y2−6x−4y+9=0 bisect the circumference of circle x2+y2−8x−6y+23=0
Similarly one can check for other options.