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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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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted June 25, 2013 at 2:29 AM via web

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Given the equation: 3x^2 + 3y^2 + 36x + 3y + 102 = 0

a) What is Center C?

b) What is the length of the radius r? 

Tagged with algebra 2, conic, math

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llltkl | College Teacher | Valedictorian

Posted June 25, 2013 at 3:28 AM (Answer #1)

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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`

`rArr (x+6)^2+(y+1/2)^2=9/4`

`rArr (x+6)^2+(y+1/2)^2=(3/2)^2`

Therefore, a) center C is (-6,-1/2) and length of the radius r is 3/2 or 1.5 units.

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