Answer :
### Problem 1: Reduce to Standard Form and Find Center and Radius
#### Equation 1: [tex]\(x^2 + y^2 + 4x - 6y - 12 = 0\)[/tex]
1. Group [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms:
[tex]\[ x^2 + 4x + y^2 - 6y = 12 \][/tex]
2. Complete the square for [tex]\( x \)[/tex]:
- Take the coefficient of [tex]\( x \)[/tex], which is 4, divide by 2 (getting 2), and square it, which gives 4.
- Add and subtract this square inside the equation.
[tex]\[ x^2 + 4x + 4 + y^2 - 6y = 12 + 4 \][/tex]
[tex]\[ (x + 2)^2 + y^2 - 6y = 16 \][/tex]
3. Complete the square for [tex]\( y \)[/tex]:
- Take the coefficient of [tex]\( y \)[/tex], which is -6, divide by 2 (getting -3), and square it, which gives 9.
- Add and subtract this square inside the equation.
[tex]\[ (x + 2)^2 + (y - 3)^2 - 9 = 16 \][/tex]
[tex]\[ (x + 2)^2 + (y - 3)^2 = 25 \][/tex]
4. Identify the center and radius:
- The center is [tex]\((-2, 3)\)[/tex].
- The radius is [tex]\(\sqrt{25} = 5\)[/tex].
So, the standard form is:
[tex]\[ (x + 2)^2 + (y - 3)^2 = 25 \][/tex]
#### Equation 2: [tex]\(x^2 + y^2 - 8x + 2y - 16 = 0\)[/tex]
1. Group [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms:
[tex]\[ x^2 - 8x + y^2 + 2y = 16 \][/tex]
2. Complete the square for [tex]\( x \)[/tex]:
- Take the coefficient of [tex]\( x \)[/tex], which is -8, divide by 2 (getting -4), and square it, which gives 16.
- Add and subtract this square inside the equation.
[tex]\[ x^2 - 8x + 16 + y^2 + 2y = 16 + 16 \][/tex]
[tex]\[ (x - 4)^2 + y^2 + 2y = 32 \][/tex]
3. Complete the square for [tex]\( y \)[/tex]:
- Take the coefficient of [tex]\( y \)[/tex], which is 2, divide by 2 (getting 1), and square it, which gives 1.
- Add and subtract this square inside the equation.
[tex]\[ (x - 4)^2 + (y + 1)^2 - 1 = 32 \][/tex]
[tex]\[ (x - 4)^2 + (y + 1)^2 = 33 \][/tex]
4. Identify the center and radius:
- The center is [tex]\((4, -1)\)[/tex].
- The radius is [tex]\(\sqrt{33}\)[/tex].
So, the standard form is:
[tex]\[ (x - 4)^2 + (y + 1)^2 = 33 \][/tex]
### Problem 2: Equation of Circle with Ends of Diameter at [tex]\((3,0)\)[/tex] and [tex]\((-2,-4)\)[/tex]
1. Find the midpoint (center of the circle):
The midpoint [tex]\( M \)[/tex] of the endpoints [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Substituting in our points [tex]\((3,0)\)[/tex] and [tex]\((-2,-4)\)[/tex]:
[tex]\[ M = \left( \frac{3 + (-2)}{2}, \frac{0 + (-4)}{2} \right) \][/tex]
[tex]\[ M = \left( \frac{1}{2}, -2 \right) \][/tex]
2. Find the radius:
The radius is half the distance between the endpoints.
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-2 - 3)^2 + (-4 - 0)^2} \][/tex]
[tex]\[ \text{Distance} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \][/tex]
The radius [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{\sqrt{41}}{2} \][/tex]
3. Write the standard form of the circle's equation:
The standard form 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.
[tex]\[ \left( x - \frac{1}{2} \right)^2 + (y + 2)^2 = \left( \frac{\sqrt{41}}{2} \right)^2 \][/tex]
Simplifying:
[tex]\[ \left( x - \frac{1}{2} \right)^2 + (y + 2)^2 = \frac{41}{4} \][/tex]
Multiplying both sides by 4 to clear the fraction:
[tex]\[ 4 \left( x - \frac{1}{2} \right)^2 + 4(y + 2)^2 = 41 \][/tex]
Expanding and simplifying:
[tex]\[ (2x - 1)^2 + 4(y + 2)^2 = 41 \][/tex]
Therefore, the equation of the circle is:
[tex]\[ (2x - 1)^2 + 4(y + 2)^2 = 41 \][/tex]
#### Equation 1: [tex]\(x^2 + y^2 + 4x - 6y - 12 = 0\)[/tex]
1. Group [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms:
[tex]\[ x^2 + 4x + y^2 - 6y = 12 \][/tex]
2. Complete the square for [tex]\( x \)[/tex]:
- Take the coefficient of [tex]\( x \)[/tex], which is 4, divide by 2 (getting 2), and square it, which gives 4.
- Add and subtract this square inside the equation.
[tex]\[ x^2 + 4x + 4 + y^2 - 6y = 12 + 4 \][/tex]
[tex]\[ (x + 2)^2 + y^2 - 6y = 16 \][/tex]
3. Complete the square for [tex]\( y \)[/tex]:
- Take the coefficient of [tex]\( y \)[/tex], which is -6, divide by 2 (getting -3), and square it, which gives 9.
- Add and subtract this square inside the equation.
[tex]\[ (x + 2)^2 + (y - 3)^2 - 9 = 16 \][/tex]
[tex]\[ (x + 2)^2 + (y - 3)^2 = 25 \][/tex]
4. Identify the center and radius:
- The center is [tex]\((-2, 3)\)[/tex].
- The radius is [tex]\(\sqrt{25} = 5\)[/tex].
So, the standard form is:
[tex]\[ (x + 2)^2 + (y - 3)^2 = 25 \][/tex]
#### Equation 2: [tex]\(x^2 + y^2 - 8x + 2y - 16 = 0\)[/tex]
1. Group [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms:
[tex]\[ x^2 - 8x + y^2 + 2y = 16 \][/tex]
2. Complete the square for [tex]\( x \)[/tex]:
- Take the coefficient of [tex]\( x \)[/tex], which is -8, divide by 2 (getting -4), and square it, which gives 16.
- Add and subtract this square inside the equation.
[tex]\[ x^2 - 8x + 16 + y^2 + 2y = 16 + 16 \][/tex]
[tex]\[ (x - 4)^2 + y^2 + 2y = 32 \][/tex]
3. Complete the square for [tex]\( y \)[/tex]:
- Take the coefficient of [tex]\( y \)[/tex], which is 2, divide by 2 (getting 1), and square it, which gives 1.
- Add and subtract this square inside the equation.
[tex]\[ (x - 4)^2 + (y + 1)^2 - 1 = 32 \][/tex]
[tex]\[ (x - 4)^2 + (y + 1)^2 = 33 \][/tex]
4. Identify the center and radius:
- The center is [tex]\((4, -1)\)[/tex].
- The radius is [tex]\(\sqrt{33}\)[/tex].
So, the standard form is:
[tex]\[ (x - 4)^2 + (y + 1)^2 = 33 \][/tex]
### Problem 2: Equation of Circle with Ends of Diameter at [tex]\((3,0)\)[/tex] and [tex]\((-2,-4)\)[/tex]
1. Find the midpoint (center of the circle):
The midpoint [tex]\( M \)[/tex] of the endpoints [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
Substituting in our points [tex]\((3,0)\)[/tex] and [tex]\((-2,-4)\)[/tex]:
[tex]\[ M = \left( \frac{3 + (-2)}{2}, \frac{0 + (-4)}{2} \right) \][/tex]
[tex]\[ M = \left( \frac{1}{2}, -2 \right) \][/tex]
2. Find the radius:
The radius is half the distance between the endpoints.
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-2 - 3)^2 + (-4 - 0)^2} \][/tex]
[tex]\[ \text{Distance} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \][/tex]
The radius [tex]\( r \)[/tex] is:
[tex]\[ r = \frac{\sqrt{41}}{2} \][/tex]
3. Write the standard form of the circle's equation:
The standard form 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.
[tex]\[ \left( x - \frac{1}{2} \right)^2 + (y + 2)^2 = \left( \frac{\sqrt{41}}{2} \right)^2 \][/tex]
Simplifying:
[tex]\[ \left( x - \frac{1}{2} \right)^2 + (y + 2)^2 = \frac{41}{4} \][/tex]
Multiplying both sides by 4 to clear the fraction:
[tex]\[ 4 \left( x - \frac{1}{2} \right)^2 + 4(y + 2)^2 = 41 \][/tex]
Expanding and simplifying:
[tex]\[ (2x - 1)^2 + 4(y + 2)^2 = 41 \][/tex]
Therefore, the equation of the circle is:
[tex]\[ (2x - 1)^2 + 4(y + 2)^2 = 41 \][/tex]
