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Answer:
Center: (3,-1)
Radius: 3
Step-by-step explanation:
Given:
[tex]\displaystyle \large{x^2+y^2-6x+2y+1=0}[/tex]
First, we have to convert the following standard circle equation to this:
[tex]\displaystyle \large{(x-h)^2+(y-k)^2=r^2}[/tex]
where h is horizontal shift, k is vertical shift and r is radius.
That means we have to complete the square for both x-term and y-term.
Rearrange the equation:
[tex]\displaystyle \large{x^2-6x+y^2+2y+1=0}\\\displaystyle \large{(x^2-6x)+(y^2+2y+1)=0}[/tex]
For [tex]\displaystyle \large{y^2+2y+1}[/tex], can be converted to perfect square as [tex]\displaystyle \large{(y+1)^2}[/tex]. Hence:
[tex]\displaystyle \large{(x^2-6x)+(y+1)^2=0}[/tex]
For the x-terms, we have to find another value that can complete the square. We know that [tex]\displaystyle \large{(a\pm b)^2 = a^2 \pm 2ab + b^2}[/tex].
For [tex]\displaystyle \large{x^2-6x}[/tex] can be [tex]\displaystyle \large{x^2-2(x)(3)+3^2 \to x^2-6x+9}[/tex]. So our another value is 9.
[tex]\displaystyle \large{(x^2-6x+9-9)+(y+1)^2=0}[/tex]
From above, we add -9 because the original expression isn’t actual perfect square.
Separate -9 out of [tex]\displaystyle \large{x^2-6x+9}[/tex]:
[tex]\displaystyle \large{(x^2-6x+9)-9+(y+1)^2=0}[/tex]
Transport -9 to add another side:
[tex]\displaystyle \large{(x^2-6x+9)+(y+1)^2=9}[/tex]
Complete the square:
[tex]\displaystyle \large{(x-3)^2+(y+1)^2=9}[/tex]
Finally, we have our needed equation to find radius and center. The coordinate of center is defined as the point (h,k) from [tex]\displaystyle \large{(x-h)^2+(y-k)^2=r^2}[/tex] and the radius is defined as r.
Hence, from the equation:
The coordinate of center is (3,-1) with radius equal to 3.