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

Question

2605 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-01T11:13:06

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

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

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

by rearranging the terms:

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

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

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

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

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

$x^{2} + 4 x + 4 + y^{2} + 8 y + 16 - 5 = 0 + 4 + 16$ $x^2+4x+4+y^2+8y+16-5=0+4+16$

$(x^{2} + 4 x + 4) + (y^{2} + 8 y + 16) = 5 + 4 + 16$ $(x^2+4x+4)+(y^2+8y+16)=5+4+16$

${(x + 2)}^{2} + {(y + 4)}^{2} = 25$ $(x+2)^2+(y+4)^2=25$

From the standard form:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

so that

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

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

How do you find the standard form of x2+y2+8y+4x−5=0x^2+y^2+8y+4x-5=0?