IDNLearn.com is the place where your questions are met with thoughtful and precise answers. Our Q&A platform is designed to provide quick and accurate answers to any questions you may have.
Sagot :
Let's solve each part step-by-step to find the center and radius of the circles.
### Part (a)
The equation of the circle is given as:
[tex]\[ x^2 + y^2 = 49 \][/tex]
The general form of a circle's equation is:
[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.
In the given equation:
- The term [tex]\(x^2 + y^2\)[/tex] indicates the circle is centered at the origin [tex]\((0, 0)\)[/tex].
- The right side of the equation, [tex]\(49\)[/tex], represents [tex]\(r^2\)[/tex].
So:
- The center [tex]\((h, k)\)[/tex] is [tex]\( (0, 0) \)[/tex]
- The radius [tex]\(r\)[/tex] is [tex]\(\sqrt{49} = 7\)[/tex]
Answer for Part (a):
- Center: [tex]\((0, 0)\)[/tex]
- Radius: [tex]\(7\)[/tex]
### Part (b)
The equation of the circle is given as:
[tex]\[ (x + 5)^2 + (y - 3)^2 = 9 \][/tex]
Comparing this with the general form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]:
- The term [tex]\((x + 5)\)[/tex] corresponds to [tex]\(x - (-5)\)[/tex], so [tex]\(h = -5\)[/tex].
- The term [tex]\((y - 3)\)[/tex] corresponds to [tex]\(y - 3\)[/tex], so [tex]\(k = 3\)[/tex].
- The right side of the equation, [tex]\(9\)[/tex], represents [tex]\(r^2\)[/tex].
So:
- The center [tex]\((h, k)\)[/tex] is [tex]\((-5, 3)\)[/tex]
- The radius [tex]\(r\)[/tex] is [tex]\(\sqrt{9} = 3\)[/tex]
Answer for Part (b):
- Center: [tex]\((-5, 3)\)[/tex]
- Radius: [tex]\(3\)[/tex]
### Part (c)
The endpoints of the diameter are given as [tex]\((-1, -1)\)[/tex] and [tex]\((-25, -11)\)[/tex].
To find the center and radius:
1. Center: The center of the circle is the midpoint of the diameter.
[tex]\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Substituting the given points:
[tex]\[ \left( \frac{-1 + (-25)}{2}, \frac{-1 + (-11)}{2} \right) = \left( \frac{-26}{2}, \frac{-12}{2} \right) = (-13, -6) \][/tex]
2. Radius: The radius is half the distance between the endpoints of the diameter.
The distance [tex]\(d\)[/tex] between points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the given points:
[tex]\[ d = \sqrt{(-25 - (-1))^2 + (-11 - (-1))^2} = \sqrt{(-24)^2 + (-10)^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \][/tex]
Radius [tex]\(r\)[/tex] is:
[tex]\[ r = \frac{26}{2} = 13 \][/tex]
Answer for Part (c):
- Center: [tex]\((-13, -6)\)[/tex]
- Radius: [tex]\(13\)[/tex]
### Part (d)
The endpoints of the diameter are given as [tex]\((4, 0)\)[/tex] and [tex]\((-2, 8)\)[/tex].
To find the center and radius:
1. Center: The center of the circle is the midpoint of the diameter.
[tex]\[ \text{Midpoint} = \left( \frac{4 + (-2)}{2}, \frac{0 + 8}{2} \right) = \left( \frac{2}{2}, \frac{8}{2} \right) = (1, 4) \][/tex]
2. Radius: The radius is half the distance between the endpoints of the diameter.
Using the distance formula:
[tex]\[ d = \sqrt{(-2 - 4)^2 + (8 - 0)^2} = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \][/tex]
Radius [tex]\(r\)[/tex] is:
[tex]\[ r = \frac{10}{2} = 5 \][/tex]
Answer for Part (d):
- Center: [tex]\((1, 4)\)[/tex]
- Radius: [tex]\(5\)[/tex]
### Part (a)
The equation of the circle is given as:
[tex]\[ x^2 + y^2 = 49 \][/tex]
The general form of a circle's equation is:
[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.
In the given equation:
- The term [tex]\(x^2 + y^2\)[/tex] indicates the circle is centered at the origin [tex]\((0, 0)\)[/tex].
- The right side of the equation, [tex]\(49\)[/tex], represents [tex]\(r^2\)[/tex].
So:
- The center [tex]\((h, k)\)[/tex] is [tex]\( (0, 0) \)[/tex]
- The radius [tex]\(r\)[/tex] is [tex]\(\sqrt{49} = 7\)[/tex]
Answer for Part (a):
- Center: [tex]\((0, 0)\)[/tex]
- Radius: [tex]\(7\)[/tex]
### Part (b)
The equation of the circle is given as:
[tex]\[ (x + 5)^2 + (y - 3)^2 = 9 \][/tex]
Comparing this with the general form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]:
- The term [tex]\((x + 5)\)[/tex] corresponds to [tex]\(x - (-5)\)[/tex], so [tex]\(h = -5\)[/tex].
- The term [tex]\((y - 3)\)[/tex] corresponds to [tex]\(y - 3\)[/tex], so [tex]\(k = 3\)[/tex].
- The right side of the equation, [tex]\(9\)[/tex], represents [tex]\(r^2\)[/tex].
So:
- The center [tex]\((h, k)\)[/tex] is [tex]\((-5, 3)\)[/tex]
- The radius [tex]\(r\)[/tex] is [tex]\(\sqrt{9} = 3\)[/tex]
Answer for Part (b):
- Center: [tex]\((-5, 3)\)[/tex]
- Radius: [tex]\(3\)[/tex]
### Part (c)
The endpoints of the diameter are given as [tex]\((-1, -1)\)[/tex] and [tex]\((-25, -11)\)[/tex].
To find the center and radius:
1. Center: The center of the circle is the midpoint of the diameter.
[tex]\[ \text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Substituting the given points:
[tex]\[ \left( \frac{-1 + (-25)}{2}, \frac{-1 + (-11)}{2} \right) = \left( \frac{-26}{2}, \frac{-12}{2} \right) = (-13, -6) \][/tex]
2. Radius: The radius is half the distance between the endpoints of the diameter.
The distance [tex]\(d\)[/tex] between points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
Substituting the given points:
[tex]\[ d = \sqrt{(-25 - (-1))^2 + (-11 - (-1))^2} = \sqrt{(-24)^2 + (-10)^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \][/tex]
Radius [tex]\(r\)[/tex] is:
[tex]\[ r = \frac{26}{2} = 13 \][/tex]
Answer for Part (c):
- Center: [tex]\((-13, -6)\)[/tex]
- Radius: [tex]\(13\)[/tex]
### Part (d)
The endpoints of the diameter are given as [tex]\((4, 0)\)[/tex] and [tex]\((-2, 8)\)[/tex].
To find the center and radius:
1. Center: The center of the circle is the midpoint of the diameter.
[tex]\[ \text{Midpoint} = \left( \frac{4 + (-2)}{2}, \frac{0 + 8}{2} \right) = \left( \frac{2}{2}, \frac{8}{2} \right) = (1, 4) \][/tex]
2. Radius: The radius is half the distance between the endpoints of the diameter.
Using the distance formula:
[tex]\[ d = \sqrt{(-2 - 4)^2 + (8 - 0)^2} = \sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \][/tex]
Radius [tex]\(r\)[/tex] is:
[tex]\[ r = \frac{10}{2} = 5 \][/tex]
Answer for Part (d):
- Center: [tex]\((1, 4)\)[/tex]
- Radius: [tex]\(5\)[/tex]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.
