Question

1. this is the equation of a circle. $x^{2}+y^{2}-2x - 4y-31 = 0$ a. complete the square and write the given equation in standard form.

Explanation:

Step1: Group x - terms and y - terms

$(x^{2}-2x)+(y^{2}-4y)=31$

Step2: Complete the square for x - terms

For $x^{2}-2x$, add $(\frac{-2}{2})^2 = 1$ to both sides of the equation.
$(x^{2}-2x + 1)+(y^{2}-4y)=31 + 1$

Step3: Complete the square for y - terms

For $y^{2}-4y$, add $(\frac{-4}{2})^2=4$ to both sides of the equation.
$(x^{2}-2x + 1)+(y^{2}-4y + 4)=31+1 + 4$

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.
$(x - 1)^{2}+(y - 2)^{2}=36$

Answer:

$(x - 1)^{2}+(y - 2)^{2}=36$