Question

How do you find the radius of a circle with the equation `x^2 + y^2 + 4x - 8y + 13 = 0`?

Answer 1

`"radius "=sqrt7`

Explanation:

`"the standard form of the equation of a circle is"`

`color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))`

`"where "(a,b)" are the coordinates of the centre and r"`
`"is the radius"`

`"to obtain this form "color(blue)"complete the square"" on both"`
`"the x and y terms"`

`x^2+4x+y^2-8y=-13`

`rArrx^2+2(2)xcolor(red)(+4)+y^2+2(-4)ycolor(magenta)(+16)=-13color(red)(+4)color(magenta)(+16)`

`rArr(x+2)^2+(y-4)^2=7larrcolor(blue)"in standard form"`

`rArrr^2=7rArrr=sqrt7`

Answer 2

`r = sqrt(7)`

Explanation:

Given: circle equation `x^2 + y^2 + 4x - 8y + 13 = 0`

The standard form of circle equation is: `(x-h)^2 + (y-k)^2 = r^2`

where center`= (h, k)`, `" "r =`radius

Use completing of the square to find the equation of the circle:

First combine the `x` terms and the `y` terms and put the constant(s) on the right side of the equation:

`(x^2 + 4x) + (y^2 - 8y) = -13`

To complete the square take `1/2` of the `x`-term `= 1/2 * 4 = 2` and `1/2` of the `y`-term `= 1/2 * -8 = -4`.

Add the extra added to the left side when you complete the square (`2^2` and `(-4)^2)` to the right side:

`(x+2)^2 + (y - 4)^2 = -13 + 2^2 + (-4)^2`

Simplify:

`(x+2)^2 + (y - 4)^2 = 7`

Since `r^2 = 7, " "r = sqrt(7)`