Question

1. -/1 points use the method of completing the square to write an equation of the circle in standard form. x² + y² - 8x - 14y + 62 = 0

Explanation:

Step1: Group x and y terms

$(x^{2}-8x)+(y^{2}-14y)+62 = 0$

Step2: Complete the square for x - terms

For $x^{2}-8x$, add and subtract $(\frac{-8}{2})^2=16$.
$(x^{2}-8x + 16-16)+(y^{2}-14y)+62 = 0$
$(x - 4)^{2}-16+(y^{2}-14y)+62 = 0$

Step3: Complete the square for y - terms

For $y^{2}-14y$, add and subtract $(\frac{-14}{2})^2 = 49$.
$(x - 4)^{2}-16+(y^{2}-14y+49 - 49)+62 = 0$
$(x - 4)^{2}-16+(y - 7)^{2}-49+62 = 0$

Step4: Simplify the equation

$(x - 4)^{2}+(y - 7)^{2}-16-49 + 62=0$
$(x - 4)^{2}+(y - 7)^{2}-3=0$
$(x - 4)^{2}+(y - 7)^{2}=3$

Answer:

$(x - 4)^{2}+(y - 7)^{2}=3$