Solve the System of Equations (x-1)^2+y^2=25 x-y^2=-4

${(x - 1)}^{2} + y^{2} = 25$ $x - y^{2} = - 4$

Step 1

Add $y^{2}$ to both sides of the equation.

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

${(x - 1)}^{2} + y^{2} = 25$

Step 2

Replace all occurrences of $x$ with $- 4 + y^{2}$ in each equation.

Replace all occurrences of $x$ in ${(x - 1)}^{2} + y^{2} = 25$ with $- 4 + y^{2}$ .

${((- 4 + y^{2}) - 1)}^{2} + y^{2} = 25$

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

Simplify the left side.

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

Simplify each term.

Subtract $1$ from $- 4$ .

${(y^{2} - 5)}^{2} + y^{2} = 25$

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

Rewrite ${(y^{2} - 5)}^{2}$ as $(y^{2} - 5) (y^{2} - 5)$ .

$(y^{2} - 5) (y^{2} - 5) + y^{2} = 25$

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

Expand $(y^{2} - 5) (y^{2} - 5)$ using the FOIL Method.

Apply the distributive property.

$y^{2} (y^{2} - 5) - 5 (y^{2} - 5) + y^{2} = 25$

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

Apply the distributive property.

$y^{2} y^{2} + y^{2} \cdot - 5 - 5 (y^{2} - 5) + y^{2} = 25$

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

Apply the distributive property.

$y^{2} y^{2} + y^{2} \cdot - 5 - 5 y^{2} - 5 \cdot - 5 + y^{2} = 25$

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

$y^{2} y^{2} + y^{2} \cdot - 5 - 5 y^{2} - 5 \cdot - 5 + y^{2} = 25$

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

Simplify and combine like terms.

Simplify each term.

Multiply $y^{2}$ by $y^{2}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y^{2 + 2} + y^{2} \cdot - 5 - 5 y^{2} - 5 \cdot - 5 + y^{2} = 25$

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

Add $2$ and $2$ .

$y^{4} + y^{2} \cdot - 5 - 5 y^{2} - 5 \cdot - 5 + y^{2} = 25$

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

$y^{4} + y^{2} \cdot - 5 - 5 y^{2} - 5 \cdot - 5 + y^{2} = 25$

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

Move $- 5$ to the left of $y^{2}$ .

$y^{4} - 5 \cdot y^{2} - 5 y^{2} - 5 \cdot - 5 + y^{2} = 25$

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

Multiply $- 5$ by $- 5$ .

$y^{4} - 5 y^{2} - 5 y^{2} + 25 + y^{2} = 25$

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

$y^{4} - 5 y^{2} - 5 y^{2} + 25 + y^{2} = 25$

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

Subtract $5 y^{2}$ from $- 5 y^{2}$ .

$y^{4} - 10 y^{2} + 25 + y^{2} = 25$

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

$y^{4} - 10 y^{2} + 25 + y^{2} = 25$

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

$y^{4} - 10 y^{2} + 25 + y^{2} = 25$

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

Add $- 10 y^{2}$ and $y^{2}$ .

$y^{4} - 9 y^{2} + 25 = 25$

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

$y^{4} - 9 y^{2} + 25 = 25$

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

$y^{4} - 9 y^{2} + 25 = 25$

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

$y^{4} - 9 y^{2} + 25 = 25$

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

Step 3

Solve for $y$ in $y^{4} - 9 y^{2} + 25 = 25$ .

Substitute $u = y^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 9 u + 25 = 25$

$u = y^{2}$

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

Subtract $25$ from both sides of the equation.

$u^{2} - 9 u + 25 - 25 = 0$

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

Combine the opposite terms in $u^{2} - 9 u + 25 - 25$ .

Subtract $25$ from $25$ .

$u^{2} - 9 u + 0 = 0$

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

Add $u^{2} - 9 u$ and $0$ .

$u^{2} - 9 u = 0$

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

$u^{2} - 9 u = 0$

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

Factor $u$ out of $u^{2} - 9 u$ .

Factor $u$ out of $u^{2}$ .

$u \cdot u - 9 u = 0$

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

Factor $u$ out of $- 9 u$ .

$u \cdot u + u \cdot - 9 = 0$

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

Factor $u$ out of $u \cdot u + u \cdot - 9$ .

$u (u - 9) = 0$

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

$u (u - 9) = 0$

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$u = 0$

$u - 9 = 0$

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

Set $u$ equal to $0$ .

$u = 0$

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

Set $u - 9$ equal to $0$ and solve for $u$ .

Set $u - 9$ equal to $0$ .

$u - 9 = 0$

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

Add $9$ to both sides of the equation.

$u = 9$

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

$u = 9$

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

The final solution is all the values that make $u (u - 9) = 0$ true.

$u = 0, 9$

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

Substitute the real value of $u = y^{2}$ back into the solved equation.

$y^{2} = 0$

$y^{2} = 9$

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

Solve the first equation for $y$ .

$y^{2} = 0$

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

Solve the equation for $y$ .

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

$y = \pm \sqrt{0}$

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

Simplify $\pm \sqrt{0}$ .

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

$y = \pm \sqrt{0^{2}}$

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

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

$y = \pm 0$

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

Plus or minus $0$ is $0$ .

$y = 0$

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

$y = 0$

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

$y = 0$

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

Solve the second equation for $y$ .

$y^{2} = 9$

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

Solve the equation for $y$ .

Remove parentheses.

$y^{2} = 9$

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

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

$y = \pm \sqrt{9}$

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

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

$y = \pm \sqrt{3^{2}}$

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

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

$y = \pm 3$

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

$y = \pm 3$

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

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.

$y = 3$

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

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

$y = - 3$

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

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

$y = 3, - 3$

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

$y = 3, - 3$

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

$y = 3, - 3$

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

The solution to $y^{4} - 9 y^{2} + 25 = 25$ is $y = 0, 3, - 3$ .

$y = 0, 3, - 3$

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

$y = 0, 3, - 3$

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

Step 4

Replace all occurrences of $y$ with $0$ in each equation.

Replace all occurrences of $y$ in $x = - 4 + y^{2}$ with $0$ .

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

$y = 0$

Simplify the right side.

Simplify $- 4 + {(0)}^{2}$ .

Raising $0$ to any positive power yields $0$ .

$x = - 4 + 0$

$y = 0$

Add $- 4$ and $0$ .

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

Step 5

Replace all occurrences of $y$ with $3$ in each equation.

Replace all occurrences of $y$ in $x = - 4 + y^{2}$ with $3$ .

$x = - 4 + {(3)}^{2}$

$y = 3$

Simplify the right side.

Simplify $- 4 + {(3)}^{2}$ .

Raise $3$ to the power of $2$ .

$x = - 4 + 9$

$y = 3$

Add $- 4$ and $9$ .

$x = 5$

$y = 3$

$x = 5$

$y = 3$

$x = 5$

$y = 3$

$x = 5$

$y = 3$

Step 6

Replace all occurrences of $y$ with $0$ in each equation.

Replace all occurrences of $y$ in $x = - 4 + y^{2}$ with $0$ .

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

$y = 0$

Simplify the right side.

Simplify $- 4 + {(0)}^{2}$ .

Raising $0$ to any positive power yields $0$ .

$x = - 4 + 0$

$y = 0$

Add $- 4$ and $0$ .

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

$x = - 4$

$y = 0$

Step 7

Replace all occurrences of $y$ with $3$ in each equation.

Replace all occurrences of $y$ in $x = - 4 + y^{2}$ with $3$ .

$x = - 4 + {(3)}^{2}$

$y = 3$

Simplify the right side.

Simplify $- 4 + {(3)}^{2}$ .

Raise $3$ to the power of $2$ .

$x = - 4 + 9$

$y = 3$

Add $- 4$ and $9$ .

$x = 5$

$y = 3$

$x = 5$

$y = 3$

$x = 5$

$y = 3$

$x = 5$

$y = 3$

Step 8

Replace all occurrences of $y$ with $- 3$ in each equation.

Replace all occurrences of $y$ in $x = - 4 + y^{2}$ with $- 3$ .

$x = - 4 + {(- 3)}^{2}$

$y = - 3$

Simplify the right side.

Simplify $- 4 + {(- 3)}^{2}$ .

Raise $- 3$ to the power of $2$ .

$x = - 4 + 9$

$y = - 3$

Add $- 4$ and $9$ .

$x = 5$

$y = - 3$

$x = 5$

$y = - 3$

$x = 5$

$y = - 3$

$x = 5$

$y = - 3$

Step 9

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

$(- 4, 0)$

$(5, 3)$

$(5, - 3)$

Step 10

The result can be shown in multiple forms.

Point Form:

$(- 4, 0), (5, 3), (5, - 3)$

Equation Form:

$x = - 4, y = 0$

$x = 5, y = 3$

$x = 5, y = - 3$

Step 11

Solve the System of Equations (x-1)^2+y^2=25 x-y^2=-4

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