Question

What is the equation of the given circle in standard form: x2 - 4x + y2 + 6y – 36 = 0?

Answer 1

Complete the squares on the given equation to write it in the standard form of the equation of a circle. Remember that the standard form of the equation of a circle with radius r and center (h,k) is:

`(x-h)^2+(y-k)^2=r^2`

Additionally, remember that to complete the square of an expression, we can add and substract the necessary constants to make it look like a perfect square trinomial:

`(z+a)^2=z^2+2az+a^2`

Starting from the given equation:

`x^2-4x+y^2+6y-36=0`

Add and substract 4 to complete the square of the x variable:

`\begin{gathered} \Rightarrow x^2-4x+4-4+y^2+6y-36=0 \\ \Rightarrow(x-2)^2-4+y^2+6y-36=0 \end{gathered}`

Add an substract 9 to complete the square of the y variable:

`\begin{gathered} \Rightarrow(x-2)^2-4+y^2+6y+9-9-36=0 \\ \Rightarrow(x-2)^2-4+(y+3)^2-9-36=0 \\ \Rightarrow(x-2)^2+(y+3)^2-9-36-4=0 \\ \Rightarrow(x-2)^2+(y+3)^2-49=0 \\ \Rightarrow(x-2)^2+(y+3)^2=49 \\ \Rightarrow(x-2)^2+(y+3)^2=7^2 \end{gathered}`

Therefore, the equation of the given circle in standard form, is:

`(x-2)^2+(y+3)^2=7^2`