Sketching Circles Example 6 - Calculus

Hi, everyone! Welcome back to integralcalc.com. Today, we’re going to be sketching the graph
of a circle. And the equation of the circle is 9x^2 + 9y^2 – 6x – 12y = 11. We’re
going to be using this formula here and what we want to do is transform the equation we
were given into the form given in this formula here. And when we’ve got it in that form,
we’re going to look at x sub 1 and y sub 1. We’ll pull out those two pieces of information
and that will give us the coordinates of the center of our circle and we’ll also grab
the radius from right here and when we have the coordinates of the center and the radius,
we’ll be able to draw our perfect circle around that center and that’ll all be the
information we need to complete our problem. So, what we need to do first in order to convert
our equation into the form given in this formula is divide through the entire equation by 9.
The reason we need to do that is because we need to make the coefficient on the x^2 and
the y^2 term equal to 1.  Right now it’s equal to 9 so we need to divide through everything
by 9 in order to get that to be 1. So when we divide through by 9, we’ll get x^2 +
y^2 – (2/3)x – (4/3)y = 11/9. So we divide everything through by 9. Then we need to group
our x terms and our y terms together. We’re not going to perform operations. We’re just
going to re-order the terms so that the x’s and y’s are grouped together. So we’ll
say [x^2 – (2/3)x] + [y^2 – (4/3)y] = 11/9. You also want to make sure when you group
your x’s and y’s together that you at the same time move your constants over to
the right-hand side. In our case, that’s already done. We’ve got 11/9 over at the
right-hand side and no other constants on the left. But if you had other constants hanging
out, you’d want to move that to the right-hand side and combine it with your 11/9 right there.
Once, we’ve got everything in this form, we need to complete the square with regards
to both x and y. And the way that we do that is standard for any ‘completing the square’
procedure. What you do is you take the coefficient on your first degree term, your second degree
term is x^2, second degree because the exponent is 2 so your second degree terms are x^2 and
y^2. Your first degree terms are raised to the powers of 1 so that’s your x term right
here and your y term right here. we want to take the coefficients on our first degree
terms. When we complete the square with respect to x, we’re going to take the coefficient
on our first degree term which is -2/3. We’re going to take that coefficient and completing
the square tells us that we need to divide that coefficient by 2. So (–2/3)/2 = -1/3.
Whatever we get for our answer, we then square it. So (-1/3)^2 = 1/9. Now, we’ll do the
same thing with respect to y. We’re going to take the coefficient on our first degree
term, -4/3. So (-4/3)/2. When we do that, we’ll get -2/3. We take that answer and
square it so (-2/3)^2 = 4/9. Now that we’ve got that, what we want to
do is add these to our equation to complete the square. We’ll get x^2 – (2/3)x, and
then we will add these to our x terms and that will complete the square. So x^2 – (2/3)x
+ 1/9 + y^2 – (4/3)y + 4/9. But then we need to remember to add 1/9 and 4/9 to the
right-hand side as well. We can’t just add them to the left-hand side. It unbalances
our equation. So we have to add 1/9 and 4/9 to the right-hand side as well. So add it
to the left, then make sure you add it to the right. Now, these completes the square
because if you factor our x group, we’ll get (x – 1/3)(x – 1/3). If we factored
the y group we’ll get (y – 2/3)(y – 2/3). And that’s equal to 16/9. The reason that
it’s called completing the square is because when you do that, you end up with perfect
squares. Notice that our factors here are the same. (x – 1/3) and (x – 1/3) are
identical. (y – 2/3) and (y – 2/3) are identical. So we can rewrite that as (x – 1/3)^2
and (y – 2/3)^2, equals 16/9. Now, you can see our equation is in the exact
same form as our formula right here so we can now take out the coordinates of the center
of our circle and our radius. So the coordinates of our center are going to be (1/3, 2/3) because
we see x sub 1 here and y sub 1 here. It’s very important to remember that in your formula,
you’ve got negative signs. If we have negative signs, we can just take the number, 1/3 and
2/3. If we had a positive sign, we would have to instead of just taking 1/3, we’d have
to take -1/3 in order to match the form in our function. But because we have negative
signs in our formula and negative signs in our factors, we can just take 1/3 and 2/3.
Our center is at the point (1/3, 2/3). Notice that our formula tells us that the radius
is going to be the square root of the right-hand side. The right-hand side is the radius squared.
In order to find just the radius on its own, we have to take the square root of the right-hand
side. And when we do that, we’ll get 4/3. You can take the square root of the numerator
and the denominator separately. The square root of 16 is 4. The square root of 9 is 3.
So our radius is equal to 4/3. Knowing that information, we can then graph
our circle. So we’ll mark off our axis in increments of 1/3. So 1/3, 2/3, 1. So our
center is at (1/3,2/3) and then our radius is 4/3. So we need to move over four tick
marks. It’s going to be a big circle. It’s going
to look like this. That’s all we do. We find the coordinates of our center and then
we move out in a distance of the radius in every direction and draw a circle around the
center. So you would draw the circle like that and then you would draw the circle and
then say the center is at (1/3,2/3) and the radius of this circle is 4/3. That’s it.
I hope this video helped you guys and I will see you in the next one. Bye!