Answer
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Work Step by Step
$x^{2}+y^{2} = 1$ Equation $(1)$
$y=x^{2}-9$ Equation $(2)$
From Equation $(2)$
$y=x^{2}-9$
$y+9=x^{2}$
$x^{2}=y+9$
Substituting $x^{2}=y+9$ in Equation $(1)$
$x^{2}+y^{2} = 1$
$y+9+y^{2} = 1$
$y^{2}+y+9-1 = 0$
$y^{2}+y+8= 0$
Using quadratic formula, $a=1, b = 1, c=8$
$y = \frac{-b±\sqrt (b^{2}-4ac)}{2a}$
$y = \frac{-1±\sqrt ((1)^{2}-4(1)(8))}{2(1)}$
$y = \frac{-1±\sqrt (1-32)}{2}$
$y = \frac{-1±\sqrt (-31)}{2}$
Since $\sqrt (-31)$ is not a real number, there is no real solution.
Solution set is empty.
