A circle has the equation x2−12x+27=−y2+8y.
What is the equation of the circle in standard form, the location of its center, and the length of its radius?

The equation of the circle is (x−6)2+(y−4)2=25; the center is at (6,4), and the radius is 5 units.

The equation of the circle is (x−6)2+(y−4)2=5; the center is at (6,4), and the radius is 5–√ units.

The equation of the circle is (x−6)2+(y−4)2=5; the center is at (−6,−4), and the radius is 5–√ units.

The equation of the circle is (x−6)2+(y−4)2=25; the center is at (−6,−4), and the radius is 5 units.

It's the first option.

Step-by-step explanation:

x^2 - 12x + 27 = - y^2 + 8y

x^2 - 12x + y^2 - 8y = -27    Completing the squares:

(x - 6)^2  - 36  + (y - 4)^2 - 16 = -27

(x - 6)^2 + (y - 4)^2 = -27 + 36 + 16

(x- 6)^2 + (y - 4)^2 = 25 is the standard equation.

The center is at (6, 4) and the radius is 5 units.

first option

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

x² - 12x + 27 = - y² + 8y ( subtract - y² + 8y from both sides )

x² - 12x + y² - 8y + 27 = 0 ( subtract 27 from both sides )

x² - 12x + y² - 8y = - 27

Using the method of completing the square on both the x and y terms

add ( half the coefficient of the x/ y term )² to both sides

x² + 2(- 6)x + 36 + y² + 2(- 4)y + 16 = - 27 + 36 + 16

(x - 6)² + (y - 4)² = 25 ← in standard form

with centre = (6, 4 ) and r² = 25 ⇒ r = = 5 → first option

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)