Answer :
To determine the equation of the circle, we start by identifying the center and the radius. Let's start by analyzing the given equation of the circle:
[tex]\[ x^2 + y^2 - 8x - 6y + 24 = 0 \][/tex]
We need to rewrite this equation in the standard form of a circle's equation:
[tex]\[ (x-h)^2 + (y-k)^2 = r^2 \][/tex]
### Step-by-Step Solution:
1. Complete the square for the [tex]\(x\)[/tex] terms:
The portion involving [tex]\(x\)[/tex] is:
[tex]\[ x^2 - 8x \][/tex]
To complete the square:
[tex]\[ x^2 - 8x = (x - 4)^2 - 16 \][/tex]
2. Complete the square for the [tex]\(y\)[/tex] terms:
The portion involving [tex]\(y\)[/tex] is:
[tex]\[ y^2 - 6y \][/tex]
To complete the square:
[tex]\[ y^2 - 6y = (y - 3)^2 - 9 \][/tex]
3. Substitute back into the original equation:
Now we substitute into the original equation, replacing the terms with their completed square forms:
[tex]\[ (x - 4)^2 - 16 + (y - 3)^2 - 9 + 24 = 0 \][/tex]
4. Simplify the equation:
Combine constants:
[tex]\[ (x - 4)^2 + (y - 3)^2 - 25 + 24 = 0 \][/tex]
[tex]\[ (x - 4)^2 + (y - 3)^2 - 1 = 0 \][/tex]
Rearrange to the standard form:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 1 \][/tex]
From this, we see that the center of the given circle is [tex]\( (4, 3) \)[/tex] and the radius is 1 unit.
5. Form the new equation:
We need to describe a new circle with the same center [tex]\( (4, 3) \)[/tex] but with a radius of 2 units (given in the problem).
The standard equation for a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
For our circle,
- center [tex]\((4, 3)\)[/tex],
- radius [tex]\(2\)[/tex],
Thus, the equation becomes:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 2^2 \][/tex]
This standard form equation of the new circle is:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 4 \][/tex]
To match the provided choices:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 2^2 \][/tex]
The correct equation is:
[tex]\[ \boxed{(x-4)^2+(y-3)^2=2^2} \][/tex]
[tex]\[ x^2 + y^2 - 8x - 6y + 24 = 0 \][/tex]
We need to rewrite this equation in the standard form of a circle's equation:
[tex]\[ (x-h)^2 + (y-k)^2 = r^2 \][/tex]
### Step-by-Step Solution:
1. Complete the square for the [tex]\(x\)[/tex] terms:
The portion involving [tex]\(x\)[/tex] is:
[tex]\[ x^2 - 8x \][/tex]
To complete the square:
[tex]\[ x^2 - 8x = (x - 4)^2 - 16 \][/tex]
2. Complete the square for the [tex]\(y\)[/tex] terms:
The portion involving [tex]\(y\)[/tex] is:
[tex]\[ y^2 - 6y \][/tex]
To complete the square:
[tex]\[ y^2 - 6y = (y - 3)^2 - 9 \][/tex]
3. Substitute back into the original equation:
Now we substitute into the original equation, replacing the terms with their completed square forms:
[tex]\[ (x - 4)^2 - 16 + (y - 3)^2 - 9 + 24 = 0 \][/tex]
4. Simplify the equation:
Combine constants:
[tex]\[ (x - 4)^2 + (y - 3)^2 - 25 + 24 = 0 \][/tex]
[tex]\[ (x - 4)^2 + (y - 3)^2 - 1 = 0 \][/tex]
Rearrange to the standard form:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 1 \][/tex]
From this, we see that the center of the given circle is [tex]\( (4, 3) \)[/tex] and the radius is 1 unit.
5. Form the new equation:
We need to describe a new circle with the same center [tex]\( (4, 3) \)[/tex] but with a radius of 2 units (given in the problem).
The standard equation for a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
For our circle,
- center [tex]\((4, 3)\)[/tex],
- radius [tex]\(2\)[/tex],
Thus, the equation becomes:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 2^2 \][/tex]
This standard form equation of the new circle is:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 4 \][/tex]
To match the provided choices:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 2^2 \][/tex]
The correct equation is:
[tex]\[ \boxed{(x-4)^2+(y-3)^2=2^2} \][/tex]
