Question

Can somebody walk me through how to solve this Algebra problem?? I'm stuck.
"Solve the following systems of a circle and a line by finding the intersection point(s), if any, algebraically.
3. y=x-4
(x+2)^2+y^2=4

Answer 1

I just graphed the 2 equations... and they don't intersect...

Answer 2

so now, it would be....
\[x ^{2}+4x+4+x ^{2}-8x+16=4\]
before i combine like terms?

Answer 3

that's the correct process... but the resulting quadratic will have complex roots.

so the lines don't intersect

Answer 4

so should i just write that??? because i have to turn in my work, but it does say "if any" lol

Answer 5

well you could do a simple sketch...

circle is centre (-2, 0) radius 2

and the line y = x - 4 has intercepts x = 4, y = -4



so all I would do is perhaps include a sketch

simply what you have to

\[x^2 - 2x + 8 = 0\]

then say there are no points on intersection

Answer 6

Thank you so much!!!

Answer 7

glad to help

Answer 8

can you help with another? I have half of it I think.... @campbell_st

Answer 9

ok... just post it

Answer 10

okay.
\[2x+y=15\]
\[(x-2)^{2}+(y-1)^{2}=25\]
I know I have to get the y on alone, so i subtracted the 2x from each side and got \[y=-2x+15\]
then i plugged that into the other equation and i guess that's where i screwed up.
@campbell_st

Answer 11

I just graphed them again, and they don't intersect...

circle has a centre at (2, 1) with radius 5

line has intercepts: x intercept 7.5 and y intercept 15

so no solutions...

perhaps before you start use this site to graph them

https://www.desmos.com/calculator

Answer 12

geez... i wonder why they're on here... Thank you though.

Answer 13

lol... its a bit odd.... but I always sketch things 1st... it may help to make sense of the solutions...