Question

The circle given by x^2+y^2-4x-10=0 can be written in standard form like this: (x-h)^2+y^2=14. What is the value of h in this equation??

Answer 1

lets rearrange it a bit

x^2-4x+y^2=10

Do you know how to complete the square?

Answer 2

No :( can you help me?

Answer 3

there should be a very very easy way to figure this out

Answer 4

you need to complete the square in x and the y's are already a perfect square

so what needs to be added to both sides to to have a perfect square in x..?

Answer 5

would you add 4x to both sides?

Answer 6

>.<

Answer 7

no?

Answer 8

x^2+y^2-4x-10=0
group x terms together and y terms together:
x^2 - 4x + y^2 - 10 = 0
complete the square for x and y separately. There is a y^2 but no y and therefore y is already a perfect square.
complete the square for x^2 - 4x
divide the coefficient of x by 2: -4/2 = -2.
The -2 will go inside the square. Subtract (-2)^2
x^2 - 4x = (x - 2)^2 - 4
y^2 = (y - 0)^2
Put them back in the original equation. Take the constant to the right.

Answer 9

would h, the constant, be 2?

Answer 10

yes. h = 2 and k = 0

Answer 11

Thank you :) for this problem do you know what the value of b would be for this circle in general form? x^2+y^2+ax+by+c=0

Answer 12

In the diagram, first locate the center of the circle and determine its coordinates.
Also find the radius.
What are they?

Answer 13

If you are having difficulty quickly locating the center of the circle you can do it the slightly long way. What is the maximum y value the circle attains? What is the minimum y value the circle attains. Add them up and divide by 2 and that will be the y coordinate of the center. Do the same with x. (Max x-value + Min. x value) / 2

Answer 14

coordinates are (1,-2) and radius is 3 correct?

Answer 15

Good job!

Now plug that into the standard form for the equation of a circle:
(x - h)^2 + (y - k)^2 = r^2 where (h, k) is the center and r is the radius.

Answer 16

(x-1)^2+(y- -2)^2=9

Answer 17

how do i know what the value of b is though?

Answer 18

We will get there. Write (y --2) as (y + 2) in the general equation first.

Answer 19

okay so (x-1)^2+(y+2)^2=9

Answer 20

Now expand the terms. Here they are interested only in b and not a and c and so it would be sufficient to expand just the (y+2)^2. But for practice expand (x-1)^2 also.

Answer 21

For example, (x-1)^2 = x^2 - 2x + 1

Answer 22

would that be (x-1)(x+1)+(y-2)(y-2)

Answer 23

yes. Or, if you remember the identity (a - b)^2 = a^2 - 2ab + b^2 you can use that too.
Or you can multiply (a+b)(a+b) if you don't know the identity.

Answer 24

so b would be 1?

Answer 25

wouldn't that be (y-2)(y+2) ?

Answer 26

No. You have to expand (y+2)^2 first.

Answer 27

(y+2)(y+2) = ?

Answer 28

y^2+4?

Answer 29

No. (y + 2)(y + 2) = y^2 + 2y + 2y + 4 = y^2 + 4y + 4

Answer 30

oh okay, so b would be 4?

Answer 31

yes.

Answer 32

thank you :)

Answer 33

Had they asked you to find a, b and c, then you will expand the whole thing:
(x-1)^2+(y+2)^2=9
x^2 - 2x + 1 + y^2 + 4y + 4 = 9
x^2 + y^2 - 2x + 4y - 4 = 0
Now you can compare it to the general form of the equation:
x^2 + y^2 + ax + by + c = 0 and conclude
a = -2; b = 4 and c = -4

Answer 34

yw.