Question

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

ITS 8!

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

Step-by-step 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.