Question

Find the center (h, k), the radius r, and the intercepts, if any of each circle.

x2 + y2 - 6x + 2y + 6 = 0

a

(h, k) = (3, -1); r = 2; x-int: 3 ± √3 (x-intercept is three plus or minus the square root of three); y-int: none


b

(h, k) = (3, -1); r = 4; x-int: 3 ± √3 (x-intercept is three plus or minus the square root of three); y-int: none

c

(h, k) = (-3, 1); r = 4; x-int: 3 ± √3 (x-intercept is three plus or minus the square root of three); y-int: none

d

(h, k) = (-3, 1); r = 2; x-int: 3 ± √3 (x-intercept is three plus or minus the square root of three); y-int: none

e

none

Answer 1

A.
`C(x,y) = (3,-1)`,
`r = 2`, x-Intercepts:
`x = 3\pm√(3)`, y-Intercepts: none.

Explanation:

Let
`x^(2)+y^(2)-6\cdot x + 2\cdot y + 6 = 0` be the general equation of the circle, we must transform the expression into standard form to determine its center, radius and intercepts. The procedure is shown below:

1)
`x^(2)+y^(2)-6\cdot x + 2\cdot y + 6 = 0` Given.

2)
`(x^(2)-6\cdot x)+(y^(2)+2\cdot y) +6= 0` Commutative and associative properties.

3)
`(x^(2)-6\cdot x)+(y^(2)+2\cdot y) + 6 + 3 + 1 = 3 + 1` Compatibility with addition.

4)
`(x^(2)-6\cdot x +9)+(y^(2)+2\cdot y +1) = 4` Definition of addition/Commutative and associative properties.

5)
`(x-3)^(2) + (y+1)^(2) = 4` Perfect square trinomial/Result.

The equation of the circle centered in (h, k) in standard form is defined as:

`(x -h)^(2) + (y-k)^(2) = r^(2)` (Eq. 1)

Where:

`h`,
`k` - Coordinate of the center of the circle, dimensionless.

`r` - Radius of the circle, dimensionless.

By direct comparison we find that circle is centered in
`C(x, y) = (3, -1)` and has a radius of 2.

Finally, we obtain the intercepts of the given function:

x-Intercepts (
`y = 0`)

`(x-3)^(2) + (0+1)^(2) = 4`

`x^(2)-6\cdot x +9+1 = 4`

`x^(2)-6\cdot x +6=0`

Roots are found analitically by Quadratic Formula:

`x = 3\pm√(3)`

y-Intercepts (
`x=0`)

`(0-3)^(2)+(y+1)^(2) = 4`

`9+y^(2)+2\cdot y +1 = 4`

`y^(2)+2\cdot y +6 = 0`

Roots are found analitically by Quadratic Formula:

`y = -1\pm i\,√(5)`

In a nutshell, there are no y-Intercepts.

We include a graphic including circle, center and x-Intercepts.

Finally, we came to the conclusion that correct answer is A.