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Ganezh [65]
3 years ago
8

Find an equation of the circle that has center (- 6, 5) and passes through (- 1, 1) .

Mathematics
1 answer:
dangina [55]3 years ago
5 0

Answer:

(x+6)^2+(y-5)^2=41

Step-by-step explanation:

The equation of a circle follows the general equation (x - h)^2 + (y - k)^2 = r^2

Since the center of the circle is (-6, 5), we progress to the formula equation (x + 6)^2 + (y- 5)^2 = r^2

Since this is a circle, the distance between (-6, 5) and (-1, 1) is the radius of the circle.

r = \sqrt{(-6 + 1)^2 + (5 - 1)^2} = \sqrt{5^2+4^2}=\sqrt{41}\\\\r^2 = 41

So we can obtain the equation (x+6)^2+(y-5)^2=41

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GET BRAINLIEST FOR THE CORRECT ANSWER AND EXPLANATION!
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Interpreting the inequality, it is found that the correct option is given by F.

------------------

  • The first equation is of the line.
  • The equal sign is present in the inequality, which means that the line is not dashed, which removes option G.

In standard form, the equation of the line is:

x + 2y = 6

2y = 6 - x

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Thus it is a decreasing line, which removes options J.

  • We are interested in the region on the plane below the line, that is, less than the line, which removes option H.

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  • As for the second equation, the normalized equation is:

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  • Thus, a circle centered at the origin and with radius 2.
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  • First, we find the following distance:

d = \sqrt{\frac{|ax + by + c|}{a^2 + b^2}}

  • Considering the coefficients of the line and the center of the circle.

d = \sqrt{\frac{|1(0) + 2(0) - 6|}{1^2 + 2^2}} = \sqrt{\frac{6}{5}} = 1.1

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A similar problem is given at brainly.com/question/16505684

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Step-by-step explanation:

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