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3. Find the center and the radius of a circle that has the equation [tex]\( x^2 + y^2 = 9 \)[/tex]
To determine the center and radius of the circle from the given equation [tex]\( x^2 + y^2 = 9 \)[/tex]:
- The equation is already in the standard form of a circle equation [tex]\( x^2 + y^2 = r^2 \)[/tex].
Here, [tex]\( r^2 = 9 \)[/tex], so the radius [tex]\( r = \sqrt{9} = 3 \)[/tex].
Since there are no terms with [tex]\( x \)[/tex] or [tex]\( y \)[/tex] shifted by any constants, the center is at the origin.
Center: [tex]\((0, 0)\)[/tex]
Radius: [tex]\(3\)[/tex]
4. Find the center and the radius of a circle that has the equation [tex]\( (x-2)^2 + (y+3)^2 = 144 \)[/tex]
To determine the center and radius of the circle from the given equation [tex]\( (x-2)^2 + (y+3)^2 = 144 \)[/tex]:
- The equation is in the form [tex]\( (x-h)^2 + (y-k)^2 = r^2 \)[/tex] where [tex]\((h, k)\)[/tex] is the center and [tex]\( r \)[/tex] is the radius.
The center [tex]\((h, k)\)[/tex] is at [tex]\( (2, -3) \)[/tex] and the radius is [tex]\( r = \sqrt{144} = 12 \)[/tex].
Center: [tex]\((2, -3)\)[/tex]
Radius: [tex]\(12\)[/tex]
5. Find the center and the radius of a circle that has the equation [tex]\( x^2 + y^2 - 6x - 12y - 55 = 0 \)[/tex]
To determine the center and radius of the circle from the given equation [tex]\( x^2 + y^2 - 6x - 12y - 55 = 0 \)[/tex]:
First, we need to complete the square to rewrite the equation in the standard form of a circle equation [tex]\( (x-h)^2 + (y-k)^2 = r^2 \)[/tex].
1. Rewrite the equation grouping [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms:
[tex]\[ x^2 - 6x + y^2 - 12y = 55 \][/tex]
2. Complete the square for the [tex]\( x \)[/tex]-terms:
[tex]\[ x^2 - 6x \quad \text{add and subtract} \left( \frac{-6}{2} \right)^2 = 9 \quad \Rightarrow \quad (x - 3)^2 - 9 \][/tex]
3. Complete the square for the [tex]\( y \)[/tex]-terms:
[tex]\[ y^2 - 12y \quad \text{add and subtract} \left( \frac{-12}{2} \right)^2 = 36 \quad \Rightarrow \quad (y - 6)^2 - 36 \][/tex]
Thus, the equation becomes:
[tex]\[ (x - 3)^2 - 9 + (y - 6)^2 - 36 = 55 \][/tex]
Simplify:
[tex]\[ (x - 3)^2 + (y - 6)^2 - 45 = 55 \quad \Rightarrow \quad (x - 3)^2 + (y - 6)^2 = 100 \][/tex]
So, the center of the circle [tex]\((h, k)\)[/tex] is at [tex]\((3, 6)\)[/tex] and the radius is [tex]\( r = \sqrt{100} = 10 \)[/tex].
Center: [tex]\((3, 6)\)[/tex]
Radius: [tex]\(10\)[/tex]
To determine the center and radius of the circle from the given equation [tex]\( x^2 + y^2 = 9 \)[/tex]:
- The equation is already in the standard form of a circle equation [tex]\( x^2 + y^2 = r^2 \)[/tex].
Here, [tex]\( r^2 = 9 \)[/tex], so the radius [tex]\( r = \sqrt{9} = 3 \)[/tex].
Since there are no terms with [tex]\( x \)[/tex] or [tex]\( y \)[/tex] shifted by any constants, the center is at the origin.
Center: [tex]\((0, 0)\)[/tex]
Radius: [tex]\(3\)[/tex]
4. Find the center and the radius of a circle that has the equation [tex]\( (x-2)^2 + (y+3)^2 = 144 \)[/tex]
To determine the center and radius of the circle from the given equation [tex]\( (x-2)^2 + (y+3)^2 = 144 \)[/tex]:
- The equation is in the form [tex]\( (x-h)^2 + (y-k)^2 = r^2 \)[/tex] where [tex]\((h, k)\)[/tex] is the center and [tex]\( r \)[/tex] is the radius.
The center [tex]\((h, k)\)[/tex] is at [tex]\( (2, -3) \)[/tex] and the radius is [tex]\( r = \sqrt{144} = 12 \)[/tex].
Center: [tex]\((2, -3)\)[/tex]
Radius: [tex]\(12\)[/tex]
5. Find the center and the radius of a circle that has the equation [tex]\( x^2 + y^2 - 6x - 12y - 55 = 0 \)[/tex]
To determine the center and radius of the circle from the given equation [tex]\( x^2 + y^2 - 6x - 12y - 55 = 0 \)[/tex]:
First, we need to complete the square to rewrite the equation in the standard form of a circle equation [tex]\( (x-h)^2 + (y-k)^2 = r^2 \)[/tex].
1. Rewrite the equation grouping [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms:
[tex]\[ x^2 - 6x + y^2 - 12y = 55 \][/tex]
2. Complete the square for the [tex]\( x \)[/tex]-terms:
[tex]\[ x^2 - 6x \quad \text{add and subtract} \left( \frac{-6}{2} \right)^2 = 9 \quad \Rightarrow \quad (x - 3)^2 - 9 \][/tex]
3. Complete the square for the [tex]\( y \)[/tex]-terms:
[tex]\[ y^2 - 12y \quad \text{add and subtract} \left( \frac{-12}{2} \right)^2 = 36 \quad \Rightarrow \quad (y - 6)^2 - 36 \][/tex]
Thus, the equation becomes:
[tex]\[ (x - 3)^2 - 9 + (y - 6)^2 - 36 = 55 \][/tex]
Simplify:
[tex]\[ (x - 3)^2 + (y - 6)^2 - 45 = 55 \quad \Rightarrow \quad (x - 3)^2 + (y - 6)^2 = 100 \][/tex]
So, the center of the circle [tex]\((h, k)\)[/tex] is at [tex]\((3, 6)\)[/tex] and the radius is [tex]\( r = \sqrt{100} = 10 \)[/tex].
Center: [tex]\((3, 6)\)[/tex]
Radius: [tex]\(10\)[/tex]
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