Question

complete the square and write the given equation in standard form. then give the center and radius of the circle and graph the equation. $x^{2}+y^{2}-4x - 6y-23 = 0$ the equation in standard form is . (simplify your answer.)

Explanation:

Step1: Group x - terms and y - terms

$(x^{2}-4x)+(y^{2}-6y)=23$

Step2: Complete the square for x - terms

For $x^{2}-4x$, add $(\frac{-4}{2})^{2}=4$ to both sides of the equation.
$(x^{2}-4x + 4)+(y^{2}-6y)=23 + 4$

Step3: Complete the square for y - terms

For $y^{2}-6y$, add $(\frac{-6}{2})^{2}=9$ to both sides of the equation.
$(x^{2}-4x + 4)+(y^{2}-6y + 9)=23+4 + 9$

Step4: Write in standard form

The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center and $r$ is the radius.
$(x - 2)^{2}+(y - 3)^{2}=36$

Answer:

$(x - 2)^{2}+(y - 3)^{2}=36$