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[tex]$x^2+y^2-4x+12y-20=0$[/tex] & [tex]$(x-6)^2+(y-4)^2=56$[/tex] \\
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[tex]$x^2+y^2+6x-8y-10=0$[/tex] & [tex]$(x-2)^2+(y+6)^2=60$[/tex] \\
\hline
[tex]$3x^2+3y^2+12x+18y-15=0$[/tex] & [tex]$(x+2)^2+(y+3)^2=18$[/tex] \\
\hline
[tex]$5x^2+5y^2-10x+20y-30=0$[/tex] & [tex]$(x+1)^2+(y-6)^2=46$[/tex] \\
\hline
[tex]$2x^2+2y^2-24x-16y-8=0$[/tex] & [tex]$x^2+y^2+2x-12y-9=0$[/tex] \\
\end{tabular}

Sagot :

GinnyAnsweridn
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]
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