The correct option is B 2(x2+y2)−2(h+2a)+ky=0
The equation of the normal at (am2,2am) is
y+mx=2am+am3
It passes through (h,k).
am3+m(2a−h)−k=0
The roots are m1,m2 and m3
∑m1=0,∑m1m2=2a−ha
m1m2m3=ka
Let equation of circle be x2+y2+2gx+2fy+C=0
It passes through (am2,2am)
a2m4+4a2m2+2agm2+4afm+C=0
a2m4+m2(4a2+2ag)+4afm+C=0
Its roots are m1,m2,m3 and m4
m1+m2+m3+m4=0
∵m1+m2+m3=0
⇒m4=0⇒ the circle passes (0,0)
m1m2+m2m3+m3m4+m4m1+m1m3+m2m4
=4a2+2aga2
⇒2a−ha=4a2+2aga2
⇒2a−h=4a+2g
⇒g=−h−2a2
m1m2m3+m2m3m4+m3m4m1+m4m1m2=−4afa2
⇒ka=−4afa2
⇒f=−k4
The equation of circle is,
x2+y2−(h+2a)x+k2y=0
⇒2(x2+y2)−2(h+2a)+ky=0