When connected in parallel
Ceq=C1+C2
When in series
Ceq′=C1+C2C1C2
C1+C2=10(C1+C2C1C2)
(C1+C2)2=10C1C2
C12+C22−8C1C2=0
dividing by C12
(C1C2)2−C18C2+1=0
Let C1C2=x
x2−8x+1=0
x=28±64−4=4+15
This question was Bonus in JEE Main 2021 exam because of incorrect data in original question.