Question

How do you write a standard form of the equation given then graphing it given 2x² + 2y² - 20x + 8y + 34 = 0 ??

Answer 1

`(x-5)^2+(y+2)^2=(sqrt12)^2` is a circle with center `(5,-2)` and radius `sqrt12`

Explanation:

This is a quadratic equation in two variables, where one - coefficients of `x^2` and `y^2` are equal and two - the term `xy` is not there (i.e. its coeffcients is `0`). Hence it represents a circle and its standarrd form is

`(x-h)^2+(y-k)^2=r^2`, where `(h,k)` is the center and radius is `r`.

Let us transform the equation `2x^2+2y^2-20x+8y+34=0` to this form. Dividing eac term by `2` - thecommon coefficient of `x^2` and `y^2` and rearranging the terms, we get

`x^2-10x+y^2+4y+17=0` and now competing sqaures

`x^2-10x+25+y^2+4y+4=25+4-17`

or `(x-5)^2+(y+2)^2=12`

i.e. `(x-5)^2+(y+2)^2=(sqrt12)^2`

and this is a circle with center `(5,-2)` and radius `sqrt12`

graph{((x-5)^2+(y+2)^2-12)((x-5)^2+(y+2)^2-0.02)=0 [-6.29, 13.71, -6.92, 3.08]}

Answer 2

The equation in standard form is `(x-5)^2+(y+2)^2=12`.

It's a circle centered at `(5,-2)` with a radius of `2sqrt3` units.

Explanation:

`2x^2+2y^2-20x+8y+34=0`

`x^2+y^2-10x+4y+17=0`

`x^2-10x+y^2+4y=-17`

`x^2-10xcolor(red)+color(red)25+y^2+4y=-17color(red)+color(red)25`

`color(blue)(x^2-10x+25)+y^2+4y=8`

`color(blue)((x-5)^2)+y^2+4y=8`

`color(blue)((x-5)^2)+y^2+4ycolor(red)+color(red)4=8color(red)+color(red)4`

`color(blue)((x-5)^2)+color(blue)(y^2+4y+4)=12`

`color(blue)((x-5)^2)+color(blue)((y+2)^2)=12`

We can see that this is an equation for a circle with center `(5,-2)` and radius `sqrt12` (or `2sqrt3`) in the form of `(x-h)^2+(y-k)^2=r^2`.

This is what its graph looks like:

graph{2x^2+2y^2-20x+8y+34=0 [-5, 15, -7, 3]}