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The equation of a circle is x2 − 4x + y2 + 8y − 29 = 0. what are the center and radius of the circle? enter your answers in the boxes.

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\text{The standard form of the circle:}\\\\(x-h)^2+(y-k)^2=r^2\\\\\text{h and k are the x and y coordinates of the circle}


\text{We have}\\\\x^2-4x+y^2+8y-29=0\\\\\text{use:(*)}(a\pm b)^2=a^2\pm2ab+b^2

x^2-2x\cdot2+y^2+2y\cdot4=29\ \ \ |+2^2\ \ |+4^2\\\\\underbrace{x^2-2x\cdot2+2^2}_((*))+\underbrace{y^2+2y\cdot4+4^2}_((*))=29+2^2+4^2\\\\(x-2)^2+(y+4)^2=49

\text{Answer: the center (2;-4), the radius}\ r=√(49)=7
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