Question

directions: select the correct location on the graph.
consider the following equation.

select the radius of the correct graph of the circle given above.

Explanation:

Step1: Complete the square for \(x\) and \(y\) terms

Given equation \(x^{2}+y^{2}-6x + 2y+6 = 0\).
Group \(x\) - terms and \(y\) - terms: \((x^{2}-6x)+(y^{2}+2y)=-6\).
For \(x\) - terms: \(x^{2}-6x=(x - 3)^{2}-9\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(b = 3\)).
For \(y\) - terms: \(y^{2}+2y=(y + 1)^{2}-1\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\), \(b = 1\)).
Substitute back: \((x - 3)^{2}-9+(y + 1)^{2}-1=-6\).

Step2: Simplify the equation

Simplify the left - hand side: \((x - 3)^{2}+(y + 1)^{2}-10=-6\).
Add \(10\) to both sides: \((x - 3)^{2}+(y + 1)^{2}=4\).

The standard form of a circle equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Here \(h = 3\), \(k=-1\), \(r^{2}=4\), so \(r = 2\).

Answer:

The radius of the circle is \(2\).