Answer :
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]
