Find the Properties 7x^2+x+7y^2+7y-1=0

Problem

7*x2+x+7*y2+7*y−1=0

Solution

1. Group the x and y terms and move the constant to the right side of the equation.

7*x2+x+7*y2+7*y=1

2. Factor out the coefficient 7 from the x and y terms to prepare for completing the square.

7*(x2+1/7*x)+7*(y2+y)=1

3. Complete the square for both variables by adding (1/2⋅1/7)2=1/196 for x and (1/2⋅1)2=1/4 for y inside the parentheses.

7*(x2+1/7*x+1/196)+7*(y2+y+1/4)=1+7*(1/196)+7*(1/4)

4. Simplify the right side and write the left side as squared binomials.

7*(x+1/14)2+7*(y+1/2)2=1+1/28+7/4

7*(x+1/14)2+7*(y+1/2)2=28/28+1/28+49/28

7*(x+1/14)2+7*(y+1/2)2=78/28

5. Divide both sides by 7 to put the equation in the standard form of a circle (x−h)2+(y−k)2=r2

(x+1/14)2+(y+1/2)2=78/196

6. Identify the center (h,k) and the radius r from the standard form.

Center=(−1/14,−1/2)

r2=78/196⇒r=√(,78)/14

Final Answer

Center: *(−1/14,−1/2), Radius: √(,78)/14

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