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

1 Answer

Answer:

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..