Question

1. in the xy - plane, the graph of ( x ^ { 2 } + y ^ { 2 } - 6 x + 2 y = - 1 ) is a circle. what is the radius of the circle?

(a) 1
(b) 3
(c) ( 2 sqrt { 3 } )
(d) 9

Explanation:

Step1: Complete the square for x

For the \(x\) - terms \(x^{2}-6x\), we use the formula \((a - b)^2=a^{2}-2ab + b^{2}\). Here \(a = x\) and \(2ab=6x\), so \(b = 3\). Then \(x^{2}-6x=(x - 3)^{2}-9\).

Step2: Complete the square for y

For the \(y\) - terms \(y^{2}+2y\), using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), with \(a = y\) and \(2ab = 2y\), so \(b = 1\). Then \(y^{2}+2y=(y + 1)^{2}-1\).

Step3: Rewrite the equation of the circle

Substitute the completed - square expressions into the original equation \(x^{2}+y^{2}-6x + 2y=-1\):

$$
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$$

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

Answer:

B. 3