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ANSWER

The radius is 8

EXPLANATION

We were given,

[tex] {x}^{2} + {y}^{2} - 2x + 8y - 47 = 0[/tex]




We rewrite the above equation to obtain,

[tex] {x}^{2} - 2x \: \: \: \: + {y}^{2} + 8y \: \: \: \: = 47[/tex]


We now add half the square of the coefficient of
[tex]x \: and \: y[/tex]
to both sides of the equation to get,


[tex] {x}^{2} - 2x + ( - 1) ^{2} + {y}^{2} + 8y + {(4)}^{2} = 47 + ( - 1) ^{2} + {4}^{2} [/tex]



We now got two perfect squares on the left hand side of the equation,


[tex] (x - 1)^{2} + {(y + 4)}^{2} = 47 + 1 + 16[/tex]


[tex] (x - 1)^{2} + {(y + 4)}^{2} =64[/tex]



[tex] (x - 1)^{2} + {(y + 4)}^{2} = {8}^{2} [/tex]



By comparing to the general formula of the circle,


[tex] {(x - a)}^{2} + {(y - b)}^{2} = {r}^{2} [/tex]


We can see that the radius is 8.

Answer:  The radius of the circle is 8 units.

Step-by-step explanation:  We are given to find the radius of a circle given by the following equation:

[tex]x^2+y^2-2x+8y-47=0~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

The standard equation of a CIRCLE with center (h, k) and radius 'r' units is given by

[tex](x-h)^2+(y-k)^2=r^2.[/tex]

From equation (i), we have

[tex]x^2+y^2-2x+8y-47=0\\\\\Rightarrow (x^2-2x+1)+(y^2+8y+16)-1-16-47=0\\\\\Rightarrow (x-1)^2+(y+4)^2-64=0\\\\\Rightarrow (x-1)^2+(y+4)^2=64\\\\\Rightarrow (x-1)^2+(y+4)^2=8^2.[/tex]

Comparing this equation with the standard equation of a circle, we get

r = 8 units.

Thus, the radius of the circle is 8 units.

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