The graph of the equation LaTeX: x^2+y^2-20x+14y+28=0 x 2 + y 2 − 20 x + 14

Question

The graph of the equation LaTeX: x^2+y^2-20x+14y+28=0
x
2
+
y
2

20
x
+
14
y
+
28
=
0
is a circle. What is the radius of this circle?
Group of answer choices

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Adalynn 3 weeks 2021-12-27T17:22:42+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-12-27T17:24:18+00:00

    Answer:

    radius = 11

    Step-by-step explanation:

    The equation of a circle in general form is

    x² + y² + 2gx + 2fy + c = 0

    with radius = \sqrt{g^2+f^2-c}

    x² + y² – 20x + 14y + 28 = 0 ← is in general form

    with 2g = – 20 ⇒ g = – 10

    2f = 14 ⇒ f = 7 and c = 28, thus

    radius = \sqrt{(-10)^2+7^2-28} = \sqrt{100+49-28} = \sqrt{121} = 11

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