Question

write the equation in standard form for the circle ( x^{2}+y^{2}-36 = 0 ).

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. For a circle centered at the origin \((h = 0,k = 0)\), the equation simplifies to \(x^{2}+y^{2}=r^{2}\).

Step2: Rearrange the given equation

Given \(x^{2}+y^{2}-36 = 0\), add \(36\) to both sides of the equation.

$$x^{2}+y^{2}-36+36=0 + 36$$

$$x^{2}+y^{2}=36$$

Answer:

\(x^{2}+y^{2}=36\)