Consider the curves, $x^2+y^2=1$ and $(x-1)^2+y^2=1.$ Points of intersection of two curves are :

$(\frac{1}{2}, ±\frac{\sqrt{3}}{2})$

The correct answer is Option (3) → $(\frac{1}{2}, ±\frac{\sqrt{3}}{2})$

$x^2+y^2=1$   ...(1)

$(x-1)^2+y^2=1$   ...(2)

comparing them

$x^2=(x-1)^2$

so $x-1=±x$

so $x=\frac{1}{2}$

at $x=\frac{1}{2}$

$(\frac{1}{2})^2+y^2=1⇒y^2=\frac{3}{4}⇒y=±\frac{\sqrt{3}}{2}$

Points of intersection $(\frac{1}{2},±\frac{\sqrt{3}}{2})$