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3 years ago

Find the center and radius of the circle defined by the equation x^2+y^2-7x+3y-4=0

Mathematics

1 answer:

scoray [572]3 years ago

3 0

Answer:

C. center: (7/2, -3/2); radius: sqrt(74)/2

Step-by-step explanation:

x^2 + y^2 - 7x + 3y - 4 = 0

We can put the equation in standard form by completing the square in x and in y.

x^2 - 7x + ___ + y^2 + 3y + ___ = 4 + ___ + ___

x^2 - 7x + (7/2)^2 + y^2 + 3y + (3/2)^2 = 4 + (7/2)^2 + (3/2)^2

(x - 7/2)^2 + (y + 3/2)^2 = 16/4 + 49/4 + 9/4

(x - 7/2)^2 + (y + 3/2)^2 = 74/4

(x - 7/2)^2 + (y + 3/2)^2 = (sqrt(74)/2)^2

Answer: center: (7/2, -3/2); radius: sqrt(74)/2

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Previous concepts

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