Answer :
To find the radius of a circle given its equation, \(x^2 + y^2 - 10x + 6y + 18 = 0\), we need to convert the equation to the standard form of a circle's equation, \((x - h)^2 + (y - k)^2 = r^2\).
Here are the steps to convert the given equation into its standard form:
1. Rewrite the given equation:
[tex]\[ x^2 + y^2 - 10x + 6y + 18 = 0 \][/tex]
2. Move the constant term to the right side of the equation:
[tex]\[ x^2 + y^2 - 10x + 6y = -18 \][/tex]
3. Group the x-terms and y-terms:
[tex]\[ (x^2 - 10x) + (y^2 + 6y) = -18 \][/tex]
4. Complete the square for the x-terms and y-terms:
For the x-terms:
[tex]\[ x^2 - 10x \][/tex]
Add and subtract \(\left(\frac{-10}{2}\right)^2 = 25\) inside the parentheses:
[tex]\[ (x^2 - 10x + 25 - 25) \][/tex]
For the y-terms:
[tex]\[ y^2 + 6y \][/tex]
Add and subtract \(\left(\frac{6}{2}\right)^2 = 9\) inside the parentheses:
[tex]\[ (y^2 + 6y + 9 - 9) \][/tex]
So, we rewrite the equation as:
[tex]\[ (x^2 - 10x + 25) + (y^2 + 6y + 9) - 25 - 9 = -18 \][/tex]
5. Simplify by combining the complete squares and constants:
[tex]\[ (x - 5)^2 + (y + 3)^2 - 34 = -18 \][/tex]
[tex]\[ (x - 5)^2 + (y + 3)^2 = 16 \][/tex]
6. Compare the resulting equation to the standard form \((x - h)^2 + (y - k)^2 = r^2\):
[tex]\[ (x - 5)^2 + (y + 3)^2 = 16 \][/tex]
Here, \( h = 5 \), \( k = -3 \), and \( r^2 = 16 \).
7. Determine the radius \(r\):
[tex]\[ r = \sqrt{16} = 4 \][/tex]
Therefore, the radius of the circle is 4 units.
Answer: 4 units.
Here are the steps to convert the given equation into its standard form:
1. Rewrite the given equation:
[tex]\[ x^2 + y^2 - 10x + 6y + 18 = 0 \][/tex]
2. Move the constant term to the right side of the equation:
[tex]\[ x^2 + y^2 - 10x + 6y = -18 \][/tex]
3. Group the x-terms and y-terms:
[tex]\[ (x^2 - 10x) + (y^2 + 6y) = -18 \][/tex]
4. Complete the square for the x-terms and y-terms:
For the x-terms:
[tex]\[ x^2 - 10x \][/tex]
Add and subtract \(\left(\frac{-10}{2}\right)^2 = 25\) inside the parentheses:
[tex]\[ (x^2 - 10x + 25 - 25) \][/tex]
For the y-terms:
[tex]\[ y^2 + 6y \][/tex]
Add and subtract \(\left(\frac{6}{2}\right)^2 = 9\) inside the parentheses:
[tex]\[ (y^2 + 6y + 9 - 9) \][/tex]
So, we rewrite the equation as:
[tex]\[ (x^2 - 10x + 25) + (y^2 + 6y + 9) - 25 - 9 = -18 \][/tex]
5. Simplify by combining the complete squares and constants:
[tex]\[ (x - 5)^2 + (y + 3)^2 - 34 = -18 \][/tex]
[tex]\[ (x - 5)^2 + (y + 3)^2 = 16 \][/tex]
6. Compare the resulting equation to the standard form \((x - h)^2 + (y - k)^2 = r^2\):
[tex]\[ (x - 5)^2 + (y + 3)^2 = 16 \][/tex]
Here, \( h = 5 \), \( k = -3 \), and \( r^2 = 16 \).
7. Determine the radius \(r\):
[tex]\[ r = \sqrt{16} = 4 \][/tex]
Therefore, the radius of the circle is 4 units.
Answer: 4 units.
