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

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Answer 1 · 2016-02-01T11:13:06

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

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

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

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..

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