Question

What is the radius of a circle given by the equation x2 + y2 – 2x + 8y – 47= 0? radius = units

Answer 1

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:

$x^2+y^2-2x+8y-47=0~~~~~~~~~~~~~~~~~~~~~(i)$

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

$(x-h)^2+(y-k)^2=r^2.$

From equation (i), we have

$x^2+y^2-2x+8y-47=0\\\\\Rightarrow (x^2-2x+1)+(y^2+8y+16)-1-16-47=0\\\\\Rightarrow (x-1)^2+(y+4)^2-64=0\\\\\Rightarrow (x-1)^2+(y+4)^2=64\\\\\Rightarrow (x-1)^2+(y+4)^2=8^2.$

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

r = 8 units.

Thus, the radius of the circle is 8 units.

Answer 2

ANSWER

The radius is 8

EXPLANATION

We were given,

${x}^{2} + {y}^{2} - 2x + 8y - 47 = 0$




We rewrite the above equation to obtain,

${x}^{2} - 2x \: \: \: \: + {y}^{2} + 8y \: \: \: \: = 47$


We now add half the square of the coefficient of
$x \: and \: y$
to both sides of the equation to get,


${x}^{2} - 2x + ( - 1) ^{2} + {y}^{2} + 8y + {(4)}^{2} = 47 + ( - 1) ^{2} + {4}^{2}$



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


$(x - 1)^{2} + {(y + 4)}^{2} = 47 + 1 + 16$


$(x - 1)^{2} + {(y + 4)}^{2} =64$



$(x - 1)^{2} + {(y + 4)}^{2} = {8}^{2}$



By comparing to the general formula of the circle,


${(x - a)}^{2} + {(y - b)}^{2} = {r}^{2}$


We can see that the radius is 8.