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1 year ago

What is the center of a circle whose equation is x2 y2 – 12x – 2y 12 = 0? (–12, –2) (–6, –1) (6, 1) (12, 2)

Mathematics

1 answer:

lora16 [44]1 year ago

8 0

The center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)

<h3>Equation of a circle</h3>

The standard equation of a circle is expressed as:

x^2 + y^2 + 2gx + 2fy + c = 0

where:

(-g, -f) is the centre of the circle

Given the equations

x^2 +y^2 – 12x – 2y +12 = 0

Compare

2gx = -12x

g = -6

Simiarly

-2y = 2fy

f = -1

Centre = (6, 1)

Hence the center of a circle whose equation is x^2 +y^2 – 12x – 2y +12 = 0 is (6,1)

Learn more on equation of a circle here: brainly.com/question/1506955

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