Question

Using the following equation, find the center and radius: x2 −2x + y2 − 6y = 26 (5 points)

Answer 1

`√(g^2+f^2-c)`

`g=-1,f=-3,c=-26`

so, the Center of the equation is
`(1,3)`

• Center → (1 , 3)

`√((-1)^2+(-3)^2-(-26))`

`=√(1+9+26)`

`=√(36)`

`=6`

• Radius → 6

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Answer 2

Center: (1,3)

Radius: 6

Explanation:

Hi there!

`x^2-2x + y^2 - 6y = 26`

Typically, the equation of a circle would be in the form
`(x-h)^2+(y-k)^2=r^2` where
`(h,k)` is the center and
`r` is the radius.

To get the given equation
`x^2-2x + y^2 - 6y = 26` into this form, we must complete the square for both x and y.

1) Complete the square for x

Let's take a look at this part of the equation:

`x^2-2x`

To complete the square, we must add to the expression the square of half of 2. That would be 1² = 1:

`x^2-2x+1`

Great! Now, let's add this to our original equation:

`x^2-2x+1+y^2-6y = 26`

We cannot randomly add a 1 to just one side, so we must do the same to the right side of the equation:

`x^2-2x+1+y^2-6y = 26+1\\x^2-2x+1+y^2-6y = 27`

Complete the square:

`(x-1)^2+y^2-6y = 27`

2) Complete the square for y

Let's take a look at this part of the equation
`(x-1)^2+y^2-6y = 27`:

`y^2-6y`

To complete the square, we must add to the expression the square of half of 6. That would be 3² = 9:

`y^2-6y+9`

Great! Now, back to our original equation:

`(x-1)^2+y^2-6y+9= 27`

Remember to add 9 on the other side as well:

`(x-1)^2+y^2-6y+9= 27+9\\(x-1)^2+y^2-6y+9= 36`

Complete the square:

`(x-1)^2+(y-3)^2= 36`

3) Determine the center and the radius

`(x-1)^2+(y-3)^2= 36`

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

Now, we can see that (1,3) is in the place of (h,k). 36 is also in the place of r², making 6 the radius.

I hope this helps!