Question

Hello, i did the problem..just needing someone else to confirm it's right. I already have the graph but just want to make sure I have the (h,k) and radius r correct incl the intercepts. THANK YOU!

1)A circle has the equation x^2+y^2+2x-2y-2=0
Graph the circle using the center (h,k) and radius r.

2) What are the intercepts, if any?

Answer 1

alright show your work

Answer 2

write in standard form
\[(x^{2} +2x) +(y^{2}-2y) = 2\]
\[(x^{2}+2x+1) +(y^{2}-2x +1) = 2+1+1\]
\[(x+1)^{2} + (y-1)^{2} = 4\]

Answer 3

from what i see you should get (h,k) = -1,1

Answer 4

and 4 is your radius ^2

Answer 5

yep

Answer 6

so your graphs should look a lot like this

Answer 7

give or take a few decimals lol

Answer 8

Intercepts, plug in zeroes
\[(0+1)^{2} + (y-1)^{2} = 4\]
\[(y-1)^{2} = 3\]
\[y = 1 \pm \sqrt{3}\]
[(x+1)^{2} + (0-1)^{2} = 4\]
\[(x+1)^{2} = 3\]
\[x = -1 \pm \sqrt{3}\]

Answer 9

A circle generally represented as\[x ^{2}+y^2+2gx+2fy+c=0\]
have centre (-g,-f)
radius \[\sqrt{g^2+f^2-c}\]
and intercepts\[(2\sqrt{g^2-c},0) and (0,2\sqrt{f^2 -c})\] if they exists

Answer 10

thank you...hello outkast, i have all my work written out...too long to type but yes, I do have that...like i mentioned to dumbcow earlier, i'm still trying to get good at getting the intercepts as it's been the biggest problem for me. but thank you so much...newbie here so it's neat to see your graph! dumbcow, too, has been such a great help and the same goes for vicky007. i have worked on 14 problems earlier today and i'm just needing 6 more (lol).... due date approaching so fast! thank you everyone.

Answer 11

your welcome and good luck with the rest

Answer 12

if you ever consider using the above formulas note these too
if g^2 -c >0 then circle cut x axis at two distinct points
if g^2-c=0 then x axis is a tangent
if g^2 -c <0 circle lies completely above or below the x axis
same applies for relation of f and c but with y axis

Answer 13

thank you...doing my best. if i can only pass this class..lol...this for sure will do it for the math part but i'm going deep into science...genetics, that is. my weakness is math, i will admit!

wow, vicky, something i would have never thought about. that's a good one too! thank you all.

Answer 14

i believe the way vicky shows is cartesian ? idk hahaha it sure looks like the cartesian