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What is the center of a circle whose equation is x2 + y2 – 12x – 2y + 12 = 0?

(–12, –2)
(–6, –1)
(6, 1)
(12, 2)

Respuesta :

xiaotongexiongp53gk1

Answer:

(6,1)

Step-by-step explanation:

Just look for the terms with x and y in them. Divide their coefficients by two, and multiply by -1. And then put them in an ordered pair.

So in this case, -12x --> 6 and -2y --> 1

Obviously the more orthodox way would be to actually complete the square, but I'm lazy, this way is faster, and since this seems to be multiple choice, as long as you have the correct answer, it shouldn't matter.

abidemiokin

The required centre will be at (6,1)

Equation of a circle

The standard equation of circle is expressed as:

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

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

Given the equation x^2 + y^2 – 12x – 2y + 12 = 0

Compare with the general to have;

2gx = -12x

g = -6

2fy = -2y

y = -1

Since centre is at (-g, -f), hence the required centre will be at (6,1)

Learn more on equation of a circle here: https://brainly.com/question/14150470

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