Answer :
Sure, let's solve the problem step by step.
### Step 1: Rewrite the Given Circle's Equation
The given equation of the circle is:
[tex]\[ 3x^2 + 3y^2 + 6x - 3y + 3 = 0 \][/tex]
First, divide the entire equation by 3 to simplify it:
[tex]\[ x^2 + y^2 + 2x - y + 1 = 0 \][/tex]
### Step 2: Convert to Standard Form
To find the center and radius, we need to rewrite the circle equation in its standard form, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 + 2x + y^2 - y = -1 \][/tex]
Complete the square for the [tex]\(x\)[/tex] terms and the [tex]\(y\)[/tex] terms:
[tex]\[ x^2 + 2x \quad \text{and} \quad y^2 - y \][/tex]
Add and subtract the necessary constant to complete the square:
For [tex]\(x\)[/tex] terms:
[tex]\[ x^2 + 2x + 1 \quad \text{(added 1 to complete the square)} \][/tex]
[tex]\[ (x + 1)^2 \][/tex]
For [tex]\(y\)[/tex] terms:
[tex]\[ y^2 - y + \frac{1}{4} \quad \text{(added } \frac{1}{4} \text{ to complete the square)} \][/tex]
[tex]\[ \left(y - \frac{1}{2}\right)^2 \][/tex]
Combine these completed squares:
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 \][/tex]
Since we added [tex]\(1 + \frac{1}{4}\)[/tex] to the left side, we must do the same to the right side:
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = -1 + 1 + \frac{1}{4} \][/tex]
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{4} \][/tex]
From this form, we see that the center of the given circle is [tex]\((-1, \frac{1}{2})\)[/tex] and its radius is [tex]\(\frac{1}{2}\)[/tex].
### Step 3: Define the New Circle's Center
A circle that is concentric with another circle has the same center. Therefore, the new circle will also have the center [tex]\((-1, \frac{1}{2})\)[/tex].
### Step 4: Calculate the Radius of the New Circle
The new circle passes through the point (1, 2). We need to find the radius of this new circle.
Using the distance formula to find the radius:
[tex]\[ r = \sqrt{(x - h)^2 + (y - k)^2} \][/tex]
[tex]\[ r = \sqrt{(1 - (-1))^2 + (2 - \frac{1}{2})^2} \][/tex]
[tex]\[ r = \sqrt{(1 + 1)^2 + \left(2 - \frac{1}{2}\right)^2} \][/tex]
[tex]\[ r = \sqrt{2^2 + \left(\frac{4}{2} - \frac{1}{2}\right)^2} \][/tex]
[tex]\[ r = \sqrt{2^2 + \left(\frac{3}{2}\right)^2} \][/tex]
[tex]\[ r = \sqrt{4 + \frac{9}{4}} \][/tex]
[tex]\[ r = \sqrt{\frac{16}{4} + \frac{9}{4}} \][/tex]
[tex]\[ r = \sqrt{\frac{25}{4}} \][/tex]
[tex]\[ r = \frac{5}{2} \][/tex]
### Step 5: Write the New Circle's Equation
The equation of the new circle in standard form is:
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{5}{2}\right)^2 \][/tex]
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{25}{4} \][/tex]
### Final Answer
Therefore, the equation of the circle concentric with [tex]\(3x^2 + 3y^2 + 6x - 3y + 3 = 0\)[/tex] and passing through the point (1, 2) is:
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{25}{4} \][/tex]
### Step 1: Rewrite the Given Circle's Equation
The given equation of the circle is:
[tex]\[ 3x^2 + 3y^2 + 6x - 3y + 3 = 0 \][/tex]
First, divide the entire equation by 3 to simplify it:
[tex]\[ x^2 + y^2 + 2x - y + 1 = 0 \][/tex]
### Step 2: Convert to Standard Form
To find the center and radius, we need to rewrite the circle equation in its standard form, [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex].
Group the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms:
[tex]\[ x^2 + 2x + y^2 - y = -1 \][/tex]
Complete the square for the [tex]\(x\)[/tex] terms and the [tex]\(y\)[/tex] terms:
[tex]\[ x^2 + 2x \quad \text{and} \quad y^2 - y \][/tex]
Add and subtract the necessary constant to complete the square:
For [tex]\(x\)[/tex] terms:
[tex]\[ x^2 + 2x + 1 \quad \text{(added 1 to complete the square)} \][/tex]
[tex]\[ (x + 1)^2 \][/tex]
For [tex]\(y\)[/tex] terms:
[tex]\[ y^2 - y + \frac{1}{4} \quad \text{(added } \frac{1}{4} \text{ to complete the square)} \][/tex]
[tex]\[ \left(y - \frac{1}{2}\right)^2 \][/tex]
Combine these completed squares:
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 \][/tex]
Since we added [tex]\(1 + \frac{1}{4}\)[/tex] to the left side, we must do the same to the right side:
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = -1 + 1 + \frac{1}{4} \][/tex]
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{4} \][/tex]
From this form, we see that the center of the given circle is [tex]\((-1, \frac{1}{2})\)[/tex] and its radius is [tex]\(\frac{1}{2}\)[/tex].
### Step 3: Define the New Circle's Center
A circle that is concentric with another circle has the same center. Therefore, the new circle will also have the center [tex]\((-1, \frac{1}{2})\)[/tex].
### Step 4: Calculate the Radius of the New Circle
The new circle passes through the point (1, 2). We need to find the radius of this new circle.
Using the distance formula to find the radius:
[tex]\[ r = \sqrt{(x - h)^2 + (y - k)^2} \][/tex]
[tex]\[ r = \sqrt{(1 - (-1))^2 + (2 - \frac{1}{2})^2} \][/tex]
[tex]\[ r = \sqrt{(1 + 1)^2 + \left(2 - \frac{1}{2}\right)^2} \][/tex]
[tex]\[ r = \sqrt{2^2 + \left(\frac{4}{2} - \frac{1}{2}\right)^2} \][/tex]
[tex]\[ r = \sqrt{2^2 + \left(\frac{3}{2}\right)^2} \][/tex]
[tex]\[ r = \sqrt{4 + \frac{9}{4}} \][/tex]
[tex]\[ r = \sqrt{\frac{16}{4} + \frac{9}{4}} \][/tex]
[tex]\[ r = \sqrt{\frac{25}{4}} \][/tex]
[tex]\[ r = \frac{5}{2} \][/tex]
### Step 5: Write the New Circle's Equation
The equation of the new circle in standard form is:
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{5}{2}\right)^2 \][/tex]
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{25}{4} \][/tex]
### Final Answer
Therefore, the equation of the circle concentric with [tex]\(3x^2 + 3y^2 + 6x - 3y + 3 = 0\)[/tex] and passing through the point (1, 2) is:
[tex]\[ (x + 1)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{25}{4} \][/tex]
