Answer
$(0,-1)$
Work Step by Step
$x^{2}+y^{2} =1$ Equation $(1)$
$x^{2}+(y+3)^{2} =4$ Equation $(2)$
From Equation $(2)$
$x^{2}+(y+3)^{2} =4$
Using $(a+b)^{2} = a^{2}+2ab+b^{2}$
$(y+3)^{2} =y^{2}+2y(3)+3^{2} = y^{2}+6y+9$
$x^{2}+ y^{2}+6y+9 =4$
From Equation $(1)$
$1+6y+9 =4$
$6y+10 =4$
$6y =4-10$
$6y =-6$
$y =-1$
Substitute $y$ value in Equation $(1)$ to get $x$
$x^{2}+y^{2} =1$
$x^{2}+(-1)^{2} =1$
$x^{2}+1 =1$
$x^{2}=0$
$x=0$
$(0,-1)$ satisfies the given equations. $(0,-1)$ is the solution.
