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Write the equation in standard form, state the center, radius, and intercepts. ...
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High School Teacher
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`
`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!
Posted by txmedteach on February 11, 2012 at 1:26 AM (Answer #1)
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