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A circle has the equation x^2+y^2+2x-2y-14=0, determine the center and radius.

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stargsd5 | Student, College Freshman | Valedictorian

Posted August 12, 2012 at 5:35 PM via web

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A circle has the equation x^2+y^2+2x-2y-14=0, determine the center and radius.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted August 12, 2012 at 5:47 PM (Answer #1)

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The equation of the circle is x^2+y^2+2x-2y-14=0. Write this in the standard form of a circle with center (h, k) and radius r: (x - h)^2 + (y - k)^2 = r^2

x^2+y^2+2x-2y-14=0

=> x^2 + 2x + 1 + y^2 - 2y + 1 = 16

=> (x + 1)^2 + (y - 1)^2 = 4^2

The center of the circle is (-1, 1) and the radius is 4.

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dlccanada | High School Teacher | eNoter

Posted August 13, 2012 at 8:20 PM (Answer #2)

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A circle has the equation x^2+y^2+2x-2y-14=0, determine the center and radius.

Let the centre of the circle (h, k) and radious is "r".The general equation of the circle

(x-h)^2 +(y-k)^2=r^2. you have to write the given equation in the general equation of the circle form

x^2+y^2+2x-2y-14=0

(x^2+2.x.1+1^2-1^2)+(y^2-2.y.1+1^2-1^1)=14

(x+1)^2-1+(y-1)^2-1=14

(x+1)^2 + (y-1)^2=14+2

(x+1)^2 + (y-1)^2=16

(x+1)^2 + (y-1)^2=4^2

there fore compare with general equation centre(-1,1) and radius is 4

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