Question

Can somebody help me write the standard form of a circle with this? `x^2+y^2-4x+8y-5=0`

Answer 1

A circle with center in `(2,-4)` and radius `5`. See below

Explanation:

Complete the squares

`x^2+y^2-4x+8y-5=(x-2)^2+(y+4)^2-4-16-5=`

`=(x-2)^2+(y+4)^2=5^2`

Answer 2

`(x-2)^2+(y--4)^2=5^2`

Explanation:

The standard Cartesian form for a circle is:

`(x-h)^2+(y-k)^2=r^2" [1]"`

Expand the squares:

`x^2-2hx+ h^2+y^2-2ky+k^2=r^2" [1.1]"`

Given:

`x^2+y^2-4x+8y-5=0" [2]"`

We need to make equation [2] look like equation [1.1], then we can make look like equation [1].

Equation [1.1] has a constant on the right. Move the -5 in equation [2] to the right by adding 5 to both sides:

`x^2+y^2-4x+8y=5" [2.1]"`

Equation [1.1] has the terms that contain x and the terms that contain y grouped together. We can rewrite equation [2.1] in the same form:

`x^2-4x+y^2+8y=5" [2.2]"`

Equation [1.1] has `h^2` and `k^2` terms. We can add those to equation [2.2] but we must add them to both sides:

`x^2-4x+ h^2+y^2+8y+ k^2=5+ h^2+k^2" [2.3]"`

You may be a bit confused because the right side has 3 terms but you will see that `5+h^2+k^2` becomes `r^2` when we find the values of h and k.

We can use the second term of equation [1.1] to find the value of h by setting it equal to the second term of equation [2.3]:

`-2hx = -4hx`

`h = 2`

We can use the fifth term of equation [1.1] to find the value of h by setting it equal to the fifth term of equation [2.3]:

`-2ky = 8y`

`k = -4`

Compute `r`:

`r^2 = 5+h^2+k^2`

`r^2 = 5+2^2+(-4)^2`

`r^2 = 25`

`r = 5`

Substitute the values of `h, k and r` into equation [1]:

`(x-2)^2+(y--4)^2=5^2`