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 l

Question

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.

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Alaia 1 month 2021-10-12T03:37:33+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-12T03:38:41+00:00

    Answer:

    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.

    0
    2021-10-12T03:38:55+00:00

    Answer:

    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 = \sqrt{25} = 5 → first option

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