Question

How do you find the standard form of `x^2+y^2+8y+4x-5=0`?

Answer 1

This is a circle `(x-(-2))^2+(y-(-4))^2=5^2`

Explanation:

From the given `x^2+y^2+8y+4x-5=0`

Perform completing the square method

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

by rearranging the terms:

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

Calculate the numbers to be added on both sides of the equation

from `4x`, take the 4, divide this number by 2 then square the result.
result `=4`

from `8y`, take the 8, divide this number by 2 then square the result.
result `=16`

Therefore Add `4` and `16` to both sides of the equation and also transpose `-5` to the right side.

`x^2+4x+4+y^2+8y+16-5=0+4+16`

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

`(x+2)^2+(y+4)^2=25`

From the standard form:

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

so that

`(x--2)^2+(y--4)^2=5^2` the required standard form.

Have a nice day !!! from the Philippines..