Question

Help me with this algebra 1 question please!

Answer 1

Center: (1, 3)

Radius: 6

Explanation:

The general equation for a circle with radius r centered at (h, k) is

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

To get the equation given into that form, we'll need to manipulate it so that the x terms and the y terms form perfect squares. That is:

`x^2-2x+a=(x+b)^2` and
`y^2-6y+c=(y+d)^2`. Let's focus on the x first. Expanding the right hand of the equation out, we have

`x^2-2x+a=x^2+2xb+b^2`

The middle term on each side tells us that

`-2x=2xb\\b=-1`

And the third term tells us that
`a=b^2=(-1)^2=1`

So, by adding a 1 to the x terms, we can "complete the square" and put them into our desired form. Let's return back to the original equation, remembering that whatever we add to one side, we have to add to the other:

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

Let's carry out the same process for the y terms. Returning to that "perfect square" equation we'd set up:

`y^2-6y+c=(y+d)^2`

Expanding the right side:

`y^2-6y+c=y^2+2yd+d^2`

Again, looking at the middle term on either side tells us that

`-6y=2yd\\-3=d\\c=d^2=(-3)^2=9`

So that perfect square equation now becomes

`y^2-6y+9=(y-3)^2`

Let's go back to our original equation on last time to add that 9 on to both sides:

`(x-1)^2+y^2-6y+9=27+9\\(x-1)^2+(y-3)^2=36`

Let's compare this to the general form for the equation of a circle:

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

We can see clearly that h = 1 and k = 3, so the center is the point (1, 3), and r² = 36, which means the radius must be √36, or 6.