If the equation y=px=a.(1+p^2)^1/2 is regarded as a quadratic in p, it will have equal root if, x^2+y^2 is
(a) -a^2
(b) 0
(c) a^2
(d) None of these
y=px=a.(1+p^2)^1/2 looks like typo (check the typing). In any case I guess this problem is about the line y=px+c being tangent to circle x2+y2=a2 for which the condtion for tangency is c2=a2(1+p2). Lines y=px±a(1+p2)1/2 would be tangents to the circle x2+y2=a2. If y=px±a(1+p2)1/2 is considered as quad. eq. in p with equal roots then these two lines would be parallel to each other.
(c).
Another way to do this is to write the eq. as (y-px)2=a2(1+p2) by squaring and then write this as quad. form in p and equate D=0.