Solve by Substitution x^2+y^2=4 , x^2+(y-1)^2=1

$x^{2} + y^{2} = 4$ , $x^{2} + {(y - 1)}^{2} = 1$

Step 1

Solve for $x$ in $x^{2} + y^{2} = 4$ .

Subtract $y^{2}$ from both sides of the equation.

$x^{2} = 4 - y^{2}$

$x^{2} + {(y - 1)}^{2} = 1$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4 - y^{2}}$

$x^{2} + {(y - 1)}^{2} = 1$

Simplify $\pm \sqrt{4 - y^{2}}$ .

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2} - y^{2}}$

$x^{2} + {(y - 1)}^{2} = 1$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = y$ .

$x = \pm \sqrt{(2 + y) (2 - y)}$

$x^{2} + {(y - 1)}^{2} = 1$

$x = \pm \sqrt{(2 + y) (2 - y)}$

$x^{2} + {(y - 1)}^{2} = 1$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{(2 + y) (2 - y)}$

$x^{2} + {(y - 1)}^{2} = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{(2 + y) (2 - y)}$

$x^{2} + {(y - 1)}^{2} = 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{(2 + y) (2 - y)}$

$x = - \sqrt{(2 + y) (2 - y)}$

$x^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$x = - \sqrt{(2 + y) (2 - y)}$

$x^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$x = - \sqrt{(2 + y) (2 - y)}$

$x^{2} + {(y - 1)}^{2} = 1$

Step 2

Solve the system $x = \sqrt{(2 + y) (2 - y)}, x^{2} + {(y - 1)}^{2} = 1$ .

Replace all occurrences of $x$ with $\sqrt{(2 + y) (2 - y)}$ in each equation.

Replace all occurrences of $x$ in $x^{2} + {(y - 1)}^{2} = 1$ with $\sqrt{(2 + y) (2 - y)}$ .

${(\sqrt{(2 + y) (2 - y)})}^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Simplify the left side.

Simplify ${(\sqrt{(2 + y) (2 - y)})}^{2} + {(y - 1)}^{2}$ .

Simplify each term.

Rewrite ${\sqrt{(2 + y) (2 - y)}}^{2}$ as $(2 + y) (2 - y)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{(2 + y) (2 - y)}$ as ${((2 + y) (2 - y))}^{\frac{1}{2}}$ .

${({((2 + y) (2 - y))}^{\frac{1}{2}})}^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${((2 + y) (2 - y))}^{\frac{1}{2} \cdot 2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Combine $\frac{1}{2}$ and $2$ .

${((2 + y) (2 - y))}^{\frac{2}{2}} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Cancel the common factor of $2$ .

Cancel the common factor.

${((2 + y) (2 - y))}^{\frac{2}{2}} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Rewrite the expression.

$((2 + y) (2 - y)) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$((2 + y) (2 - y)) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Simplify.

$(2 + y) (2 - y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$(2 + y) (2 - y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Expand $(2 + y) (2 - y)$ using the FOIL Method.

Apply the distributive property.

$2 (2 - y) + y (2 - y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Apply the distributive property.

$2 \cdot 2 + 2 (- y) + y (2 - y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Apply the distributive property.

$2 \cdot 2 + 2 (- y) + y \cdot 2 + y (- y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$2 \cdot 2 + 2 (- y) + y \cdot 2 + y (- y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $2$ .

$4 + 2 (- y) + y \cdot 2 + y (- y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Multiply $- 1$ by $2$ .

$4 - 2 y + y \cdot 2 + y (- y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Move $2$ to the left of $y$ .

$4 - 2 y + 2 \cdot y + y (- y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Rewrite using the commutative property of multiplication.

$4 - 2 y + 2 y - y \cdot y + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$4 - 2 y + 2 y - (y \cdot y) + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Multiply $y$ by $y$ .

$4 - 2 y + 2 y - y^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$4 - 2 y + 2 y - y^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$4 - 2 y + 2 y - y^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Add $- 2 y$ and $2 y$ .

$4 + 0 - y^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Add $4$ and $0$ .

$4 - y^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$4 - y^{2} + {(y - 1)}^{2} = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Rewrite ${(y - 1)}^{2}$ as $(y - 1) (y - 1)$ .

$4 - y^{2} + (y - 1) (y - 1) = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Expand $(y - 1) (y - 1)$ using the FOIL Method.

Apply the distributive property.

$4 - y^{2} + y (y - 1) - 1 (y - 1) = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Apply the distributive property.

$4 - y^{2} + y \cdot y + y \cdot - 1 - 1 (y - 1) = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Apply the distributive property.

$4 - y^{2} + y \cdot y + y \cdot - 1 - 1 y - 1 \cdot - 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$4 - y^{2} + y \cdot y + y \cdot - 1 - 1 y - 1 \cdot - 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$4 - y^{2} + y^{2} + y \cdot - 1 - 1 y - 1 \cdot - 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Move $- 1$ to the left of $y$ .

$4 - y^{2} + y^{2} - 1 \cdot y - 1 y - 1 \cdot - 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Rewrite $- 1 y$ as $- y$ .

$4 - y^{2} + y^{2} - y - 1 y - 1 \cdot - 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Rewrite $- 1 y$ as $- y$ .

$4 - y^{2} + y^{2} - y - y - 1 \cdot - 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Multiply $- 1$ by $- 1$ .

$4 - y^{2} + y^{2} - y - y + 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$4 - y^{2} + y^{2} - y - y + 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Subtract $y$ from $- y$ .

$4 - y^{2} + y^{2} - 2 y + 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$4 - y^{2} + y^{2} - 2 y + 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$4 - y^{2} + y^{2} - 2 y + 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Simplify by adding terms.

Combine the opposite terms in $4 - y^{2} + y^{2} - 2 y + 1$ .

Add $- y^{2}$ and $y^{2}$ .

$4 + 0 - 2 y + 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Add $4$ and $0$ .

$4 - 2 y + 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$4 - 2 y + 1 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Add $4$ and $1$ .

$- 2 y + 5 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$- 2 y + 5 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$- 2 y + 5 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$- 2 y + 5 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

$- 2 y + 5 = 1$

$x = \sqrt{(2 + y) (2 - y)}$

Solve for $y$ in $- 2 y + 5 = 1$ .

Move all terms not containing $y$ to the right side of the equation.

Subtract $5$ from both sides of the equation.

$- 2 y = 1 - 5$

$x = \sqrt{(2 + y) (2 - y)}$

Subtract $5$ from $1$ .

$- 2 y = - 4$

$x = \sqrt{(2 + y) (2 - y)}$

$- 2 y = - 4$

$x = \sqrt{(2 + y) (2 - y)}$

Divide each term in $- 2 y = - 4$ by $- 2$ and simplify.

Divide each term in $- 2 y = - 4$ by $- 2$ .

$\frac{- 2 y}{- 2} = \frac{- 4}{- 2}$

$x = \sqrt{(2 + y) (2 - y)}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y}{- 2} = \frac{- 4}{- 2}$

$x = \sqrt{(2 + y) (2 - y)}$

Divide $y$ by $1$ .

$y = \frac{- 4}{- 2}$

$x = \sqrt{(2 + y) (2 - y)}$

$y = \frac{- 4}{- 2}$

$x = \sqrt{(2 + y) (2 - y)}$

$y = \frac{- 4}{- 2}$

$x = \sqrt{(2 + y) (2 - y)}$

Simplify the right side.

Divide $- 4$ by $- 2$ .

$y = 2$

$x = \sqrt{(2 + y) (2 - y)}$

$y = 2$

$x = \sqrt{(2 + y) (2 - y)}$

$y = 2$

$x = \sqrt{(2 + y) (2 - y)}$

$y = 2$

$x = \sqrt{(2 + y) (2 - y)}$

Replace all occurrences of $y$ with $2$ in each equation.

Replace all occurrences of $y$ in $x = \sqrt{(2 + y) (2 - y)}$ with $2$ .

$x = \sqrt{(2 + 2) (2 - (2))}$

$y = 2$

Simplify $x = \sqrt{(2 + 2) (2 - (2))}$ .

Simplify the left side.

Remove parentheses.

$x = \sqrt{(2 + 2) (2 - (2))}$

$y = 2$

$x = \sqrt{(2 + 2) (2 - (2))}$

$y = 2$

Simplify the right side.

Simplify $\sqrt{(2 + 2) (2 - (2))}$ .

Add $2$ and $2$ .

$x = \sqrt{4 (2 - (2))}$

$y = 2$

Multiply $- 1$ by $2$ .

$x = \sqrt{4 (2 - 2)}$

$y = 2$

Subtract $2$ from $2$ .

$x = \sqrt{4 \cdot 0}$

$y = 2$

Multiply $4$ by $0$ .

$x = \sqrt{0}$

$y = 2$

Rewrite $0$ as $0^{2}$ .

$x = \sqrt{0^{2}}$

$y = 2$

Pull terms out from under the radical, assuming positive real numbers.

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

$x = 0$

$y = 2$

Step 3

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(0, 2)$

Step 4

The result can be shown in multiple forms.

Point Form:

$(0, 2)$

Equation Form:

$x = 0, y = 2$

Step 5

Solve by Substitution x^2+y^2=4 , x^2+(y-1)^2=1

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