Question

X2 + y2 +3x-6y+9=0 How can you find the center, radius, and intercepts? I'm not sure how to put in general or standard form.

Answer 1

See below.

Explanation:

`x^2+y^2+3x-6y+9=0`

We need to get this into the form:

`color(blue)((x-h)^2+(y-k)^2=r^2`

Where:

`(h,k)=>(x,y)` are the x and y coordinates of the centre receptively and `r` is the radius.

To reach this form, we complete the square on one side of the equation for both x and y.

Group terms containing x together and group terms containing y together:

`x^2+3x+y^2-6y+9`

Place a bracket around the x terms:

`(x^2+3x)+y^2-6y+9`

Working inside the bracket, add the square of half the coefficient of the x term:

`(x^2+3x+(3/2)^2)+y^2-6y+9`

Now subtract the square of half the coefficient of x outside the bracket. This is equivalent to doing nothing:

`(x^2+3x+(3/2)^2)-(3/2)^2+y^2-6y+9`

Now convert `(x^2+3x+(3/2)^2)` into the square of a binomial:

`(x+3/2)^2-(3/2)^2+y^2-6y+9`

Now do the same with the two terms containing y:

`(x+3/2)^2-(3/2)^2+(y^2-6y)+9=0`

`(x+3/2)^2-(3/2)^2+(y^2-6y+(-3)^2)-(-3)^2+9=0`

`(x+3/2)^2-(3/2)^2+(y-3)^2-(-3)^2+9=0`

Now simplify:

`(x+3/2)^2-9/4+(y-3)^2=0`

`(x+3/2)^2+(y-3)^2=9/4`

From this we see that:

`h=-3/2`

`k=3`

`r=sqrt(9/4)=3/2`

( note the signs ) This is because we had `(x-h)` and `(y-k)`

So:

`(x-(-3/2))=(x+3/2)`

`(y-(3))=(y-3)`

So the centre is at `color(blue)((-3/2 , 3 )`

And the radius is `color(blue)(3/2)`

Intercepts

y intercepts occur when `x=0`:

`((0)+3/2)^2+(y-3)^2=9/4`

`9/4+(y-3)^2=9/4`

`(y-3)^2=0`

`y-3=0=>y=3`

Only one value indicates that the circle only touches the y axis it dose not cross it. So no y intercepts

x axis intercepts occur when `y=0`:

`(x+3/2)^2+((0)-3)^2=9/4`

`(x+3/2)^2+9=9/4`

`x^2+3x+9/4+9=9/4`

`x^2+3x+9=0`

Quadratic discriminant:

`sqrt(b^2-4ac)`

`sqrt((3)^2-(4(1)(9)))=sqrt(9-36)=sqrt(-27)`

No real solutions, so no x axis intercepts

GRAPH: