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`