Question

Find the center and radius x^2 -18x +53 = 18y - y^2

Answer 1

@pitamar do you know this?

Answer 2

You're having here an equation of an ellipse. In order to solve it you first have to convert it to this form:
$$ (ax-h)^2 + (by-v)^2 = r^2 $$

Answer 3

we know that: \((a + b)^2 = a^2 + 2ab + b^2\)
So we can tell:
$$
x^2 - 18x + 53 = x^2 + 2 \cdot (-9)x + 53 = \\
= x^2 + 2 \cdot (-9)x + (-9)^2 - (-9)^2 + 53 = \\
=(x - 9)^2 - 81 + 53 = (x-9)^2 - 28
$$

Answer 4

Can you do the same for the \(y\) on the right side?

Answer 5

That whole equation just confused me

Answer 6

I though the variables for a circle are h and k

Answer 7

hehe, lol don't worry about that too much. If you're dealing with circles and not ellipse then you can rid off the \(a,b\) as well and you have:
$$(x-h)^2 + (y-v)^2 = r^2$$

Answer 8

in our case we have \((x-9)^2\) so \(h=9\)

Answer 9

That looks better Lol

Answer 10

ye I guess =p
Ok, so can you complete the square of y on the right as well?

Answer 11

Wait how did you get 9 for h

Answer 12

on the equation we have \((x-h)^2\) and so far we have \((x-9)^2\)
Don't worry about that yet, I just wanted to show you that we're getting the equation to this shape.
Now you have to complete the square for y too and we can solve it =)

Answer 13

Would y=9 too?

Answer 14

I mean k

Answer 15

ye seems so, but it's not enough, because in order to complete the square you have to add & subtract something, and that will remain in the equation afterwards.
You can see that for doing it for the x I had to add \( (-9)^2 - (-9)^2 \) in order to form it into a square and then I was left with that \(-(-9)^2\)

Answer 16

So would you need to get x and y on the left side of the equation

Answer 17

You can move the y's to the left first if you want and then complete the square, that's fine

Answer 18

You can also complete the square the way it is now on the right, whatever you prefer

Answer 19

What is missing on the right in order to complete a square for y?

Answer 20

Oh a number with a variable?

Answer 21

without

Answer 22

Ye, some number
Let me write what we have and move the y's to the left it'll be easier:
$$(x-9)^2 - 28 = 18y - y^2 \\
(x-9)^2 + y^2 - 18y = 28
$$

Answer 23

We can add a number to both sides. what number is missing in order for us to be able to form a square?

Answer 24

Oh I see what you mean now Lol. You need to subtract the 28 over and make it equal (x−9)2+y2−18y -28 = 0

Answer 25

make it equal 0 forget the giberish

Answer 26

Well the calculation is true but we don't want to get it equal to 0
We want to complete the square. so we get \((y + \text{something})^2\) instead of the \(y^2 - 18y\) we have now

Answer 27

Just remember that if we had a square then it would be of the kind \((y + n)^2 = y^2 + 2ny + n^2\)
We have to find what \(n\) is in our case in order to add \(n^2\) to both sides and then we'll be able to form the square \((y + n)^2\)

Answer 28

want a hint?

Answer 29

Yea that would help Lol I'm so confused

Answer 30

Ok ok, We can see that we have \(y^2 - 18y\) we can form it as \(y^2 + 2 (-9)y\) which reveals us that the third term has to be \((-9)^2\) in order to complete the square to \((y+ (-9))^2 = (y-9)^2\)
So we add \((-9)^2 = 81\) to both sides:
$$ (x-9)^2 + y^2 -18y = 28 \\
(x-9)^2 + y^2 -18y + 81 = 28 + 81 \\
(x-9)^2 + (y-9)^2 = 109
$$

Answer 31

makes sense?

Answer 32

Oh yea now it does xD Thanks for the help bro. You saved me

Answer 33

lol no problem ,glad I could help =)