Page 801? Numbers 32 and 36?

Here's 20

Center (1,2)   Point (0,6)   So....we can find the radius using the distance formula

r = √[(1-0)^2  + (6-2)^2] = √[(1) + 4^2] = √[1 + 16] = √17

So...we have

(x-h)^2 + (y-k)^2 = r^2     where (h,k) is the center  and r is the radius

(x-1)^2 + (y-2)^2  = 17

Nope, but it wasn't that hard to go to page 801.

for 32, the center of the circle is at (p,q) = (0,0) and the radius 3

Since the general equation of a circle is given by

$$(x-p)^2+(y-q)^2 = r^2$$ where (p,q) is the center and r the radius

the equation of this circle is

$$(x-0)^2 + (y-0)^2 = 3^2 \Rightarrow x^2+ y^2 = 9$$

Similarly for 36 we find

(p,q) = (-1,1) and r = 2

Therefore

$$(x--1)^2 + (y-1)^2 = 2^2 \Rightarrow (x+1)^2 + (y-1)^2 = 4$$

Reinout 

thank you very much Reinout! sorry for the link!

That's all right,

I wouldn't have known how else you could've referred me to that page.

What you Mean?

I know I Tried to Copy and Paste the Link to Test it Out...And it Referred Me to the Correct link, You Just Had to Type In the Number?

Maybe i Copied it wrong.

Yeah now that you say it; 

you wrote

http://www.coursesmart.com/9781256493303/429?portal=coursesmart&CSTenantKey=ashworthcollege&key=144F70C86CC50AE01C576D8C488C8E5B020C26F7ACF55E437EE2666BA1F60603486872B77CEED537672A1D4F563689C20C5FD975492123FE52A5C63690F50A5AF24FB40684C3D6DBC2873EA37A43FFA50E43F7EA555BA850F663EBEC23049ED10EE95B0332ED3989BA96F3C73D136A7F0508BEE7E70DC913833CCF7FDD15626A&spid=#X2ludGVybmFsX0J2ZGVwRmxhc2hSZWFkZXI/eG1saWQ9OTc4MTI1NjQ5MzMwMy84MDE

and it should've been

http://www.coursesmart.com/9781256493303/801?portal=coursesmart&CSTenantKey=ashworthcollege&key=144F70C86CC50AE01C576D8C488C8E5B020C26F7ACF55E437EE2666BA1F60603486872B77CEED537672A1D4F563689C20C5FD975492123FE52A5C63690F50A5AF24FB40684C3D6DBC2873EA37A43FFA50E43F7EA555BA850F663EBEC23049ED10EE95B0332ED3989BA96F3C73D136A7F0508BEE7E70DC913833CCF7FDD15626A&spid=#X2ludGVybmFsX0J2ZGVwRmxhc2hSZWFkZXI/eG1saWQ9OTc4MTI1NjQ5MzMwMy84MDE

don't sweat it though

Oh, How odd!

I Clicked on that link and It Said Session Unavailable?

I wont Sweat it :)

Have a Greate Day!

Haha yeah I had that too, but if you try again it'll send you through.

You have a great day too 

If You Aren't too Busy Can You Show me #20 On the Same Page?

I Would Be very Happy With Any More Help!

Thank You So Much CPhill!

I Needed That,

Very good Explaining too!

Common Core

yuck.