The circle $2x^2+2y^2+8x+12y-136=0$ has centre and radius:

1. Divide through by 2 so the coefficients of $x^2$ and $y^2$ are 1: $x^2+y^2+4x+6y-68=0$. 2. For $x^2+y^2+2gx+2fy+c=0$ the centre is $(-g,-f)$, here $(-2,-3)$. 3. The radius is $\sqrt{g^2+f^2-c}=\sqrt{4+9+68}=\sqrt{81}=9$. 4. Centre $(-4,-6)$ comes from reading $8$ and $12$ off the original equation without first dividing by 2. 5. Radius $81$ stops before taking the square root. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/1, Question 8: Coordinate Geometry I_