![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{6}{ h},\stackrel{-1}{ k})\qquad \qquad radius=\stackrel{4}{ r} \\\\[-0.35em] ~\dotfill\\[1em] [x-6]^2+[y-(-1)]^2=4^2\implies (x-6)^2+(y+1)^2=16 \\\\\\ \stackrel{\mathbb{F~O~I~L}}{(x^2-12x+36)}+\stackrel{\mathbb{F~O~I~L}}{(y^2+2y+1)}=16\implies x^2+y^2-12x+2y+37=16 \\\\\\ x^2+y^2-12x+2y+37-16=0\implies x^2+y^2-12x+2y+21=0](/tpl/images/0048/5786/57a07.png)
The center of a circle is located at (6, −1) . the radius of the circle is 4.
what is the equa...
Mathematics, 04.07.2019 02:20 Blazingangelkl
The center of a circle is located at (6, −1) . the radius of the circle is 4.
what is the equation of the circle in general form?
x2+y2−12x+2y+21=0
x2+y2−12x+2y+33=0
x2+y2+12x−2y+21=0
x2+y2+12x−2y+33=0
Answers: 1
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