Equation of circle in standard form is [tex] (x-h)^{2} + (y-k)^{2} =r^{2} [/tex]
where (h,k) is the center of the circle and r is the radius.
To write the given equation in standard form we take 52 to the other side .The equation we have is :
[tex] x^{2} +10x +y^{2} + 12y =-52 [/tex]
To complete the square we take half of coefficient of x and y terms then square it .This is added to both sides of the equation .Half the coefficient of x term is 5 and half the coefficient of y term is 6 .Squaring it we have [tex] 5^{2} and 6^{2} [/tex]
Adding this to both sides:
[tex] x^{2} +10x + 5^{2} +y^{2} +12y + 6 ^{2} = -52 +5^{2} + 6^{2} [/tex]
Simplifying we have:
[tex] (x+5)^{2} +(y+6) ^{2} =9 [/tex]
The last option is the right answer.
