Question

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

Answer 1

Answer:

Center: (1,3)

Radius: 6

Step-by-step 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!

Answer 2

Answer:

$\sqrt{g^2+f^2-c}$

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

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

• Center → (1 , 3)

$\sqrt{(-1)^2+(-3)^2-(-26})$

$=\sqrt{1+9+26}$

$=\sqrt{36}$

$=6$

• Radius → 6

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