Question

Please write ${x}^{2} + {y}^{2} + 6x - 14y - 42 = 0$in standard form.

Answer 1

Given the standard equation of a circle

$(x-a)^2+(y-b)^2=r^2$$\text{Where a and b are the center of the circle}$

Given the equation

$x^2+y^2+6x-14y-42=0$

Collect like terms and simplify in their complete form

$\begin{gathered} x^2+6x+y^2-14y=42 \\ (x^2+6x)+(y^2-14y)=42 \\ (x+3)^2-9+(y-7)^2-49=42 \\ (x+3)^2+(y-7)^2=42+49+9 \\ (x+3)^2+(y-7)^2=100 \\ (x+3)^2+(y-7)^2=10^2 \end{gathered}$

Hence the standard form of the equation is

$(x+3)^2+(y-7)^2=10^2$