Answer :
Sure, let's break down the steps to solve this problem.
### Step 1: Find the Center of the Given Circle
First, let's work on the given equation of the circle:
[tex]\[ x^2 + y^2 - 8x - 6y + 24 = 0 \][/tex]
To identify the center, we'll complete the square for \(x\) and \(y\).
#### Completing the Square for \(x\):
1. Extract the \(x\) terms:
[tex]\[ x^2 - 8x \][/tex]
2. Complete the square:
Add and subtract \((\frac{8}{2})^2 = 16\):
[tex]\[ x^2 - 8x + 16 - 16 \][/tex]
This simplifies to:
[tex]\[ (x - 4)^2 - 16 \][/tex]
#### Completing the Square for \(y\):
1. Extract the \(y\) terms:
[tex]\[ y^2 - 6y \][/tex]
2. Complete the square:
Add and subtract \((\frac{6}{2})^2 = 9\):
[tex]\[ y^2 - 6y + 9 - 9 \][/tex]
This simplifies to:
[tex]\[ (y - 3)^2 - 9 \][/tex]
#### Combining everything:
So, the given equation becomes:
[tex]\[ (x - 4)^2 - 16 + (y - 3)^2 - 9 + 24 = 0 \][/tex]
Simplify this:
[tex]\[ (x - 4)^2 + (y - 3)^2 - 1 = 0 \][/tex]
Thus,
[tex]\[ (x - 4)^2 + (y - 3)^2 = 1 \][/tex]
From this equation, we can see that the center \((h, k)\) of the original circle is \((4, 3)\).
### Step 2: Form the Equation for the New Circle
The problem states that the new circle has a radius of 2 units and the same center as the original circle \((4, 3)\).
The general form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\).
Substituting in the values:
- Center \((h, k) = (4, 3)\)
- Radius \(r = 2\)
The equation becomes:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 2^2 \][/tex]
### Step 3: Compare with the Given Choices
We need to match \((x - 4)^2 + (y - 3)^2 = 2^2\) with one of the provided options:
- \((x+4)^2+(y+3)^2=2\)
- \((x-4)^2+(y-3)^2=2\)
- \((x-4)^2+(y-3)^2=2^2\)
- \((x+4)^2+(y+3)^2=2^2\)
### Step 4: Verify the Correct Choice
The correct choice is the one that matches our derived equation \((x - 4)^2 + (y - 3)^2 = 2^2\):
Therefore, the correct option is:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 2^2 \][/tex]
Hence, the equation representing the new circle is:
[tex]\[ \boxed{(x-4)^2 + (y-3)^2 = 2^2} \][/tex]
### Step 1: Find the Center of the Given Circle
First, let's work on the given equation of the circle:
[tex]\[ x^2 + y^2 - 8x - 6y + 24 = 0 \][/tex]
To identify the center, we'll complete the square for \(x\) and \(y\).
#### Completing the Square for \(x\):
1. Extract the \(x\) terms:
[tex]\[ x^2 - 8x \][/tex]
2. Complete the square:
Add and subtract \((\frac{8}{2})^2 = 16\):
[tex]\[ x^2 - 8x + 16 - 16 \][/tex]
This simplifies to:
[tex]\[ (x - 4)^2 - 16 \][/tex]
#### Completing the Square for \(y\):
1. Extract the \(y\) terms:
[tex]\[ y^2 - 6y \][/tex]
2. Complete the square:
Add and subtract \((\frac{6}{2})^2 = 9\):
[tex]\[ y^2 - 6y + 9 - 9 \][/tex]
This simplifies to:
[tex]\[ (y - 3)^2 - 9 \][/tex]
#### Combining everything:
So, the given equation becomes:
[tex]\[ (x - 4)^2 - 16 + (y - 3)^2 - 9 + 24 = 0 \][/tex]
Simplify this:
[tex]\[ (x - 4)^2 + (y - 3)^2 - 1 = 0 \][/tex]
Thus,
[tex]\[ (x - 4)^2 + (y - 3)^2 = 1 \][/tex]
From this equation, we can see that the center \((h, k)\) of the original circle is \((4, 3)\).
### Step 2: Form the Equation for the New Circle
The problem states that the new circle has a radius of 2 units and the same center as the original circle \((4, 3)\).
The general form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\).
Substituting in the values:
- Center \((h, k) = (4, 3)\)
- Radius \(r = 2\)
The equation becomes:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 2^2 \][/tex]
### Step 3: Compare with the Given Choices
We need to match \((x - 4)^2 + (y - 3)^2 = 2^2\) with one of the provided options:
- \((x+4)^2+(y+3)^2=2\)
- \((x-4)^2+(y-3)^2=2\)
- \((x-4)^2+(y-3)^2=2^2\)
- \((x+4)^2+(y+3)^2=2^2\)
### Step 4: Verify the Correct Choice
The correct choice is the one that matches our derived equation \((x - 4)^2 + (y - 3)^2 = 2^2\):
Therefore, the correct option is:
[tex]\[ (x - 4)^2 + (y - 3)^2 = 2^2 \][/tex]
Hence, the equation representing the new circle is:
[tex]\[ \boxed{(x-4)^2 + (y-3)^2 = 2^2} \][/tex]
