Question

Which of the following equations listed below has a radius of 3?

O A) x^2 + y^2 + 12x - 20y+ 132 = 0
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O B) x^2 + y^2 + 12x - 20y + 55 = 0
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O C) x^2 + y^2 - 12x+ 20y + 127 = 0
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O D) x^2 + y^2 – 20x – 16y + 155 = 0

Answer 1

The answer is C I hope this helps

Answer 2

C.

Explanation:

WE need to convert the given equations into the standard form

(x - a)^2 + (y - b)^2 = r^2 where r = the radius.

x^2 + y^2 + 12x - 20y+ 132 = 0

(x + 6)^2 - 36 + (y - 10)^2 - 100 = -132

(x + 6)^2 + (y - 10)^2 = -132 + 36 + 100 = 4

giving a radius of 2

x^2 + y^2 + 12x - 20y + 55 = 0

will give a right hand side of -55 + 36 + 100 = 81 giving radius 9.

x^2 + y^2 - 12x+ 20y + 127 = 0

- gives right hand side of -127 + 36 + 100 = 9 giving a radius of 3.