Consider the curves, $x^2+y^2=1$ and $(x-1)^2+y^2=1.$ Points of intersection of two curves are :
Answer & explanation
Correct answer: option 3
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 $(\frac{1}{2},±\frac{\sqrt{3}}{2})$