Question 660817
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Hi, there--

Problem:
Find the center (h, k) and the radius r of the circle
{{{3x^2+7x+3y^2-7y-5=0}}}

Answer:
We want to rearrange this equation into the form (x-h)^2 +(y-k)^2 = r^2. Then we can read the center point and the radius right off the equation.

Divide every term in the equation by 3, so that the x-squared and y-squared terms have a coefficient of 1.

{{{x^2+(7/3)x+y^2-(7/3)y-5/3=0}}}

Add 5/3 to both sides of the equation.
{{{x^2 + (7/3)x + y^2-(7/3)y = 5/3}}}

Complete the square for the x terms by adding a constant. To make a perfect square trinomial find half the coefficient of the x-term and square it. Half of 7/3 is 7/6. Th square of 7/6 is 49/36. Add this to both sides of the equation.
{{{(x^2+(7/3)x+49/36) + y^2 -(7/3)y=5/3+49/36}}}

Repeat this process for the y-terms.
{{{(x^2+(7/3)x+49/36)+(y^2-(7/3)+49/36)=5/3+49/36+49/36}}}

Write each trinomial in factored form. Sum up the constants on the right side.
{{{(x+7/6)^2+(y-7/6)^2=158/36}}}

The square root of 158/36 is {{{(sqrt(158))/6}}} because the square root of 36 is 6. THe final equation is 

{{{(x+7/6)^2+(y-7/6)^2=((sqrt(158))/6)^2}}}

The center of the circle is the point (h,k)=(-7/6, 7/6)
The radius of the circle is {{{(sqrt(158))/6}}}

Hope this helps! Feel free to email if my explanation is unclear or you have additional questions.

Mrs.Figgy
math.in.the.vortex@gmail.com
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