The radius of a circle given by the equation [tex]\rm x^2+y^2-2x+8y-47=0[/tex] is 8 units.
It is given that the circle equation is given by [tex]\rm x^2+y^2-2x+8y-47=0[/tex]
It is required to find the radius of the circle.
It is defined as the combination of points that and every point has an equal distance from a fixed point ( called the center of a circle).
We circle the equation:
[tex]\rm x^2+y^2-2x+8y-47=0[/tex]
Converting this equation into the standard form of the circle:
[tex]\rm (x-a)^2+(y-b)^2=r^2[/tex]
[tex]\rm x^2+y^2-2x+8y-47=0[/tex]
[tex]\rm x^2+y^2-2x+8y-47+1-1+16-16=0[/tex] (by adding and subtracting by
1 and 16)
[tex]\rm x^2-2x+1+y^2+8y+16-1-16-47=0\\\\\rm (x-1)^2+(y-(-4))^2-64=0\\\\\rm (x-1)^2+(y-(-4))^2=64\\\\\rm (x-2)^2+(y-(-4))^2=8^2 \\\\[/tex]
By comparing with the standard circle equation, we get:
r = 8
Thus, the radius of a circle given by the equation [tex]\rm x^2+y^2-2x+8y-47=0[/tex] is 8 units
Learn more about circle here:
brainly.com/question/11833983
