Certainly! Let's break down each part of the question step-by-step.
### Example 5: For each of the circle, state the center and radius
#### Circle (a): [tex]\[ x^2 + y^2 = 64 \][/tex]
This equation represents a circle in standard 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.
1. Comparing [tex]\( x^2 + y^2 = 64 \)[/tex] with the standard form, we see that:
- The center is [tex]\((0, 0)\)[/tex].
- The radius [tex]\( r \)[/tex] is [tex]\(\sqrt{64} = 8.0\)[/tex].
Thus, the center is [tex]\((0, 0)\)[/tex] and the radius is [tex]\(8.0\)[/tex].
#### Circle (b): [tex]\[ x^2 + y^2 = 20 \][/tex]
This equation also represents a circle in standard form [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex].
1. Comparing [tex]\( x^2 + y^2 = 20 \)[/tex] with the standard form:
- The center is [tex]\((0, 0)\)[/tex].
- The radius [tex]\( r \)[/tex] is [tex]\(\sqrt{20} \approx 4.4721\)[/tex].
Thus, the center is [tex]\((0, 0)\)[/tex] and the radius is approximately [tex]\(4.4721\)[/tex].
#### Circle (c): [tex]\[ (x + 2)^2 + (y - 4)^2 = 81 \][/tex]
This equation again matches the standard form [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex].
1. Comparing [tex]\( (x + 2)^2 + (y - 4)^2 = 81 \)[/tex] with the standard form:
- The center is [tex]\((-2, 4)\)[/tex] (since [tex]\(x + 2\)[/tex] implies a shift left by 2 units and [tex]\(y - 4\)[/tex] implies a shift up by 4 units).
- The radius [tex]\( r \)[/tex] is [tex]\(\sqrt{81} = 9.0\)[/tex].
Thus, the center is [tex]\((-2, 4)\)[/tex] and the radius is [tex]\(9.0\)[/tex].
### Example 6: Find the center and the length of the radius of the circle and sketch it
[tex]\[ x^2 + y^2 - 6x + 4y - 30 = 0 \][/tex]
To convert this general equation into the standard form, we need to complete the square.
1. Group the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms:
[tex]\[ x^2 - 6x + y^2 + 4y = 30 \][/tex]
2. Complete the square for [tex]\( x \)[/tex]-terms and [tex]\( y \)[/tex]-terms separately:
- For [tex]\( x \)[/tex]-terms:
[tex]\[ x^2 - 6x \][/tex]
To complete the square, add and subtract [tex]\((\frac{-6}{2})^2 = 9\)[/tex]:
[tex]\[ x^2 - 6x + 9 - 9 \][/tex]
This becomes:
[tex]\[ (x - 3)^2 - 9 \][/tex]
- For [tex]\( y \)[/tex]-terms:
[tex]\[ y^2 + 4y \][/tex]
To complete the square, add and subtract [tex]\((\frac{4}{2})^2 = 4\)[/tex]:
[tex]\[ y^2 + 4y + 4 - 4 \][/tex]
This becomes:
[tex]\[ (y + 2)^2 - 4 \][/tex]
3. Substitute back into the equation:
[tex]\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 = 30 \][/tex]
Simplify:
[tex]\[ (x - 3)^2 + (y + 2)^2 - 13 = 30 \][/tex]
Add 13 to both sides:
[tex]\[ (x - 3)^2 + (y + 2)^2 = 43 \][/tex]
The standard form of the circle is [tex]\( (x - 3)^2 + (y + 2)^2 = 43 \)[/tex].
1. The center is [tex]\((3, -2)\)[/tex] (Since [tex]\(x - 3\)[/tex] implies a shift right by 3 units, and [tex]\(y + 2\)[/tex] implies a shift down by 2 units).
2. The radius [tex]\( r \)[/tex] is [tex]\(\sqrt{43} \approx 6.5574\)[/tex].
Thus, the center is [tex]\((3, -2)\)[/tex] and the radius is approximately [tex]\(6.5574\)[/tex].
