Question

https://gyazo.com/b998b374b47b49b80eb6c6ef2108d282

Answer 1

Looks like a bunch of equations of circles.
You'd get a lot of practice by putting each one into standard form for the equation of a circle centered at (h,k) and with radius r.\[x^2+y^2-2x+2y-1=0\] should be rearranged as follows, in prep for "completing the squares."

Answer 2

Group all the x terms together; then group all the y terms together.\[x^2-2x +y^2+2y-1=0\]

Answer 3

To what extent is "completing the square" familiar to you?

That x^2 -2x can be transformed into a perfect square (of a binomial), less a constant:\[x^2-2x +1 -1\rightarrow (x-1)^2-1\]

Answer 4

excuse me, \[x^2-2x+1-1 \rightarrow (x-1)^2 -1\]

Answer 5

Then, \[x^2-2x +y^2+2y-1=0\] becomes \[x^2-2x +1-1+y^2+2y-1=0, or\]

Answer 6

\[(x-1)^2-1+y^2+2y-1=0\]

Answer 7

Lot of steps! If you need help in understanding "completing the square," please say so.

the next result stems from completing the square in the case of y^2 + 2y:
\[(x-1)^2-1+[(y+1)^2 -1=0\] which simplifies to \[(x-1)^2+(y+1)^2=2=r^2\]

Answer 8

So, the length of the radius, r, is 1.

You'll have to go thru the same steps to put all of the other equations of circles into "general form" (which looks like my last result, above), and then rearrange the equations in ascending order: smallest radius first, at top of list; largest radius last, at bottom of list.

Answer 9

Since\[(x-1)^2+(y+1)^2=2=r^2,\]\[r^2=2, \]

Answer 10

and thus \[r=\sqrt{2}\]

Answer 11

@Lock sorry to go on and on without any feedback from you. I need to hear from you, as do other OpenStudy users. What's clear here? What's not?