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Write the equation in standard form, state the center, radius, and intercepts. ...

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nanalisa | Student, Undergraduate | Honors

Posted February 11, 2012 at 12:08 AM via web

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Write the equation in standard form, state the center, radius, and intercepts.  x²+y²-8x-6y=0

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted February 11, 2012 at 1:26 AM (Answer #1)

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So, to answer this question, you must be familiar with the concept of "completing the square." It is based on the fact that perfect square trinomials are of the form:

`x^2 + 2bx + b^2 = (x+b)^2`

So, to find our standard form of the circle,

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

we need to complete the square twice on the equation you provide. Let's first separate our equation in a way that will let us do this:

`x^2 - 8x + 0 + y^2 - 6y + 0 = 0`

I just added the 0's for placeholders. Now, let's start by completing the square for the x-term. Based on the form I showed you above, to find the number that will get you a perfect-square trinomial of x will be found by the following steps:

1) Divide the coefficient of `x` by 2

2) Square this number to find the constant term

So in our case, the steps look like this:

1) `-8/2 = -4`

2) `(-4)^2 = 16`

So, to make the x terms into a perfect square trinomial, we need to add 16. Notice, though that we need to add 16 to both sides of the equation!

`x^2-8x+16 + y^2 -6x + 0 = 16`

We do the same for y:

1) `-6/2 = -3`

2) `(-3)^2 = 9`

So, we'll add 9 to both sides, too! Shown here:

`x^2 - 8x + 16 + y^2 -6x + 9 = 16 + 9`

Now, we need to factor. If you'll notice based on the form I showed above, that each equation will now factor easily by taking the value from Step 1 of completing the square (remember, -4 and -3):

`(x-4)^2 + (y-3)^2 = 25`

Now, to get it in standard form, we need to find what number's square makes 25 to get our radius:

`(x-4)^2 + (y-3)^2 = 5^2`

And now, our equation is in standard form.

So, our radius and center drop out pretty easily from this:

Center = `(h,k) = (4,3)`

Radius = `sqrt(r^2) = sqrt(5^2) = 5`

Alright, now we need to find the intercepts. To do this, we'll need to remember what finding the intercept means.

At the x-intercepts, remember that for a line to cross the x-axis, the y-value of those points are zero. So, to find the x-intercepts of our graph, we must simply set y to zero and solve for x:
`(x-4)^2 + (0-3)^2 = 25`

`x^2 - 8x + 16 + 9 = 25`

Simplifying:

`x^2 - 8x + 25 = 25`

Let's subtract 25 from both sides:

`x^2 - 8x = 0`

Now to find the two roots of this polynomial, let's just factor:

`x(x-8) = 0`

So, we get the following two x-intercepts:

`x = 0, x = 8`

and therefore: `(0,0)` and `(8,0)`

Now, we need to find the y-intercepts in pretty much the same way. Just as we found the x-intercepts by setting y to zero, we find the y-intercepts by setting the x to zero:

`(0-4)^2 + (y-3)^2 = 25`

`16 + y^2 - 6y +9 = 25`

`y^2 -6y + 25 = 25`

Again, let's subtract 25 from both sides:

`y^2-6y = 0`

And to find the y intercepts, let's just factor;
`y(y-6) = 0`

So our y-values will be `y=0, y=6`, so our y-intercepts will be:

`(0,0)` and `(0,6)`

And there you have it! Let's graph it to make sure we have everything right:

Yup, looks like our numbers our right. Don't worry about the circle, it's just crushed because of the scale.

I hope that helps!

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