Let's consider the given equation pair:
[tex]\[ 2x^2 + 2y^2 - 24x - 16y - 8 = 0 \][/tex]
[tex]\[ x^2 + y^2 + 2x - 12y - 9 = 0 \][/tex]
We need to identify the center and radius for each of these equations, as well as the points of intersection.
### Step-by-Step Solution
#### Transform the First Equation
First, let’s transform the first equation to the standard form of a circle, which is [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex].
[tex]\[ 2x^2 + 2y^2 - 24x - 16y - 8 = 0 \][/tex]
Divide the entire equation by 2:
[tex]\[ x^2 + y^2 - 12x - 8y - 4 = 0 \][/tex]
Complete the square for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ (x^2 - 12x) + (y^2 - 8y) = 4 \][/tex]
For [tex]\(x\)[/tex]:
[tex]\[ x^2 - 12x = (x-6)^2 - 36 \][/tex]
For [tex]\(y\)[/tex]:
[tex]\[ y^2 - 8y = (y-4)^2 - 16 \][/tex]
Including these in our equation, we get:
[tex]\[ (x-6)^2 - 36 + (y-4)^2 - 16 = 4 \][/tex]
[tex]\[ (x-6)^2 + (y-4)^2 - 52 = 4 \][/tex]
Thus:
[tex]\[ (x-6)^2 + (y-4)^2 = 56 \][/tex]
So, the center of the circle is [tex]\((6, 4)\)[/tex] and the radius is [tex]\(\sqrt{56} \approx 7.48\)[/tex].
#### Transform the Second Equation
Next, let’s transform the second equation to the standard form of a circle.
[tex]\[ x^2 + y^2 + 2x - 12y - 9 = 0 \][/tex]
Complete the square for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ (x^2 + 2x) + (y^2 - 12y) = 9 \][/tex]
For [tex]\(x\)[/tex]:
[tex]\[ x^2 + 2x = (x+1)^2 - 1 \][/tex]
For [tex]\(y\)[/tex]:
[tex]\[ y^2 - 12y = (y-6)^2 - 36 \][/tex]
Including these in our equation, we get:
[tex]\[ (x+1)^2 - 1 + (y-6)^2 - 36 = 9 \][/tex]
[tex]\[ (x+1)^2 + (y-6)^2 - 37 = 9 \][/tex]
Thus:
[tex]\[ (x+1)^2 + (y-6)^2 = 46 \][/tex]
So, the center of the circle is [tex]\((-1, 6)\)[/tex] and the radius is [tex]\(\sqrt{46} \approx 6.78\)[/tex].
### Final Answer
The transformations result in the following data:
1. For the first circle:
- Center: [tex]\((6, 4)\)[/tex]
- Radius: [tex]\(\sqrt{56} \approx 7.48\)[/tex]
2. For the second circle:
- Center: [tex]\((-1, 6)\)[/tex]
- Radius: [tex]\(\sqrt{46} \approx 6.78\)[/tex]
