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Given the equation: 3x^2 + 3y^2 + 36x + 3y + 102 = 0 a) What is Center C? b) What is...
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The standard form of the equation of a circle is: `(x-h)^2+(y-k)^2=r^2` , where the center is (h,k) and the radius is r.
Here, the given equation is:
`3x^2 + 3y^2 + 36x + 3y + 102 = 0`
Dividing by 3,
`x^2 + y^2 + 12x + y + 34 = 0`
Grouping the x terms and y terms and rearranging:
`x^2 + 12x + y^2 + y = -34`
Now, to complete the square we have to take half of the coefficients of x and y and square them and add them to both sides of the equation.
`x^2 + 12x+(12/2)^2 + y^2 + y+(1/2)^2 = -34+(12/2)^2+(1/2)^2`
`rArr x^2 + 12x+36 + y^2 + y+1/4 = -34+36+1/4`
Therefore, a) center C is (-6,-1/2) and length of the radius r is 3/2 or 1.5 units.
Posted by llltkl on June 25, 2013 at 3:28 AM (Answer #1)
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