How do you graph $4 x^{2} + 4 y^{2} + 24 x - 32 y + 51$ $4x^2 + 4y^2 + 24x - 32y + 51$ ?

Question

2751 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-09T00:25:32

You need to complete the square for both the x and y variables

$4 x^{2} + 4 y^{2} + 24 x - 32 y + 51 = 0$ $4x^2 + 4y^2 + 24x - 32y + 51=0$

$4 (x^{2} + 6 x) + 4 (y^{2} - 8 y) = - 51$ $4(x^2 + 6x) +4(y^2 - 8y) =- 51$

Now, complete the squares remembering to keep the equation balanced :

$4 (x^{2} + 6 x + 9) + 4 (y^{2} - 8 y + 16) = - 51 + (4) (9) + (4) (16)$ $4(x^2 + 6x+ 9) +4(y^2 - 8y+16) =- 51+(4)(9)+(4)(16)$

$4 {(x + 3)}^{2} + 4 {(y - 4)}^{2} = 49$ $4(x+3)^2 +4(y-4)^2=49$

${(x + 3)}^{2} + {(y - 4)}^{2} = \frac{49}{4}$ $(x+3)^2 +(y-4)^2=49/4$

So, this is a circle with:

Center $= (- 3, 4)$ $=(-3,4)$
Radius $= \sqrt{\frac{49}{4}} = 3.5$ $=sqrt(49/4)=3.5$

Hope that helped enter image source here

How do you graph 4x2+4y2+24x−32y+514x^2 + 4y^2 + 24x - 32y + 51?