Math / Geometry

Question1. What is the equation, center, and radius of the circle? $\begin{array}{l} x^{2}+2 x+y^{2}+6 y-7=0 \\ x^{2}+2 x+y^{2}+6 y=7 \\ x^{2}+2 x+17 y^{2}+6 y=7+1 \\ (x+1)^{2}+y^{2}+6 y=8 \\ (x+1)^{2}+y^{2}+6 y+9<8+9 \\ (x+1)^{2}+(y+3)^{2}=17 \end{array}$

Studdy Solution

STEP 1

1. The given equation is a quadratic equation representing a circle.
2. The equation needs to be rewritten in the standard form of a circle's equation to identify the center and radius.

STEP 2

1. Rearrange the given equation.
2. Complete the square for both $x$ and $y$ .
3. Rewrite the equation in the standard form of a circle.
4. Identify the center and radius from the standard form.

STEP 3

Rearrange the given equation to group $x$ and $y$ terms:
$x^2 + 2x + y^2 + 6y = 7$

STEP 4

Complete the square for the $x$ terms:
$x^2 + 2x$
To complete the square, take half of the coefficient of $x$ , square it, and add it inside the equation:
$\left(\frac{2}{2}\right)^2 = 1$
Add and subtract 1:
$(x^2 + 2x + 1) - 1$
This becomes:
$(x + 1)^2 - 1$

STEP 5

Complete the square for the $y$ terms:
$y^2 + 6y$
To complete the square, take half of the coefficient of $y$ , square it, and add it inside the equation:
$\left(\frac{6}{2}\right)^2 = 9$
Add and subtract 9:
$(y^2 + 6y + 9) - 9$
This becomes:
$(y + 3)^2 - 9$

STEP 6

Rewrite the equation in the standard form of a circle:
Substitute the completed squares back into the equation:
$(x + 1)^2 - 1 + (y + 3)^2 - 9 = 7$
Simplify:
$(x + 1)^2 + (y + 3)^2 = 17$

STEP 7

Identify the center and radius from the standard form:
The standard form of a circle is:
$(x - h)^2 + (y - k)^2 = r^2$
From the equation:
$(x + 1)^2 + (y + 3)^2 = 17$
The center $(h, k)$ is $(-1, -3)$ and the radius $r$ is $\sqrt{17}$ .
The equation of the circle is:
$(x + 1)^2 + (y + 3)^2 = 17$
The center of the circle is:
$(-1, -3)$
The radius of the circle is:
$\sqrt{17}$

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Question1. What is the equation, center, and radius of the circle? $\begin{array}{l} x^{2}+2 x+y^{2}+6 y-7=0 \\ x^{2}+2 x+y^{2}+6 y=7 \\ x^{2}+2 x+17 y^{2}+6 y=7+1 \\ (x+1)^{2}+y^{2}+6 y=8 \\ (x+1)^{2}+y^{2}+6 y+9<8+9 \\ (x+1)^{2}+(y+3)^{2}=17 \end{array}$

Studdy Solution

STEP 1

1. The given equation is a quadratic equation representing a circle.
2. The equation needs to be rewritten in the standard form of a circle's equation to identify the center and radius.

STEP 2

1. Rearrange the given equation.
2. Complete the square for both $x$ and $y$ .
3. Rewrite the equation in the standard form of a circle.
4. Identify the center and radius from the standard form.

STEP 3

Rearrange the given equation to group $x$ and $y$ terms:
$x^2 + 2x + y^2 + 6y = 7$

STEP 4

Complete the square for the $x$ terms:
$x^2 + 2x$
To complete the square, take half of the coefficient of $x$ , square it, and add it inside the equation:
$\left(\frac{2}{2}\right)^2 = 1$
Add and subtract 1:
$(x^2 + 2x + 1) - 1$
This becomes:
$(x + 1)^2 - 1$

STEP 5

Complete the square for the $y$ terms:
$y^2 + 6y$
To complete the square, take half of the coefficient of $y$ , square it, and add it inside the equation:
$\left(\frac{6}{2}\right)^2 = 9$
Add and subtract 9:
$(y^2 + 6y + 9) - 9$
This becomes:
$(y + 3)^2 - 9$

STEP 6

Rewrite the equation in the standard form of a circle:
Substitute the completed squares back into the equation:
$(x + 1)^2 - 1 + (y + 3)^2 - 9 = 7$
Simplify:
$(x + 1)^2 + (y + 3)^2 = 17$

STEP 7

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Studdy Solution

STEP 1

1. The given equation is a quadratic equation representing a circle.2. The equation needs to be rewritten in the standard form of a circle's equation to identify the center and radius.

STEP 2

1. Rearrange the given equation.2. Complete the square for both xxx and yyy.3. Rewrite the equation in the standard form of a circle.4. Identify the center and radius from the standard form.

STEP 3

Rearrange the given equation to group xxx and yyy terms:x2+2x+y2+6y=7 x^2 + 2x + y^2 + 6y = 7 x2+2x+y2+6y=7

STEP 4

STEP 5

STEP 6

Rewrite the equation in the standard form of a circle:Substitute the completed squares back into the equation:(x+1)2−1+(y+3)2−9=7 (x + 1)^2 - 1 + (y + 3)^2 - 9 = 7 (x+1)2−1+(y+3)2−9=7Simplify:(x+1)2+(y+3)2=17 (x + 1)^2 + (y + 3)^2 = 17 (x+1)2+(y+3)2=17

STEP 7

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The given equation is a quadratic equation representing a circle.
2. The equation needs to be rewritten in the standard form of a circle's equation to identify the center and radius.

1. Rearrange the given equation.
2. Complete the square for both $x$ and $y$ .
3. Rewrite the equation in the standard form of a circle.
4. Identify the center and radius from the standard form.

Rearrange the given equation to group $x$ and $y$ terms:
$x^2 + 2x + y^2 + 6y = 7$

Rewrite the equation in the standard form of a circle:
Substitute the completed squares back into the equation:
$(x + 1)^2 - 1 + (y + 3)^2 - 9 = 7$
Simplify:
$(x + 1)^2 + (y + 3)^2 = 17$