Question

HELP ME! Does anyone know a simple way of solving systems involving circles?

Answer 1

Post your question and we'll see

Answer 2

x^2+y^2=36
y=2x+15

Answer 3

So I guess substitution is not the easy way?

Answer 4

x^2 + (2x+15)^2 = 36

Answer 5

Well it is but once it comes to factoring it takes a little longer then i thought

Answer 6

Well there is another way. Let's see if it works.

Answer 7

I got 5x^2+60x+189. Now i have to factor but it isn't coming out right. And i am trying the quadratic formula.

Answer 8

square the second equation to get

y^2 = (2x+15)^2
y^2 = 36 - x^2


y^2 = (2x+15)(2x+15)
y^2 = 2x(2x+15)+15(2x+15)
y^2 = 4x^2 + 30x + 30x + 225
y^2 = 4x^2 + 60x + 225


y^2 = y^2

4x^2 + 60x + 225 = 36 - x^2

5x^2 + 60x + 189 = 0

ac = 189*5
a + c = 60

That leads to a dead end unfortunately

Answer 9

I'm convinced that the roots are complex

Answer 10

They are and I dont know to go around that so i am lost.

Answer 11

There's no way around complex roots.

Answer 12

True. But for the answer Do I just put zero?

Answer 13

The answer to this depends. What exactly is the question that you were required to answer. Tell me word for word.

Answer 14

All it says is solve the systems and it just have a list of different problems for me to do.

Answer 15

Have you been studying complex roots at all?

Answer 16

If it doesn't say, you'll have to assume it wants the complex roots

Answer 17

Yes I master that. Lol. But for solving systems involving circles is very complex to understand.

Answer 18

If you can find the roots like you say you can, what exactly is the difficulty?

Answer 19

Maybe it just annoys you

Answer 20

Because I am questioning Do my teacher wants complex numbers part of my answer.

Answer 21

It kind is because i been stuck on this for a while now.

Answer 22

I'm pretty sure that's what your teacher wants. You have no choice but to provide them since he or she didn't specify.

Answer 23

True so that means my answer will be complex for the quadratic formula.

Answer 24

In this case, you have to provide the (x,y) points

Answer 25

Yes because that part of my solution.

Answer 26

There's no easy way to calculate the complex roots. As far as I'm concerned, you have three options:

1. Quad Formula
2. Complete The Square
3. Myininaya's Method

Trust me. You don't want Myininaya's Method.

Answer 27

Ok. Then thanks for your help if i still need help i just hop back online.

Answer 28

lol; it said "false" on my calculator. I should have investigated further why it said that.

Answer 29

@telliott99 I drew it out also and i got that. Now what would be my (x,y)

Answer 30

http://mrpilarski.wordpress.com/2009/12/04/solving-systems-of-equations-using-the-elimination-method/

Answer 31

I must confess, that I haven't read all the posts here with equal intensity, but if I combine what @Hero and @telliott99 suggested already, it seems like these two functions never meet.

One is a line, the other is a circle.

The solution is that there is no solution, therefore the line never intersects the circle.

Answer 32

So in case your textbook says different @TammyFlow, you might want to check the question again and see if there wasn't any confusion of \(\pm \) signs.

Answer 33

I got that by the picture i drew. so thanks @Spacelimbus . And to @TheViper I know how to do that already but thanks for the help

Answer 34

yw dude :)

Answer 35

It wasn't a textbook. It was just a sheet with problems on there about solving systems involving circles . So i am just going to draw it out for now on

Answer 36

@TheViper I am a girl not a dude.

Answer 37

LOL so wht should I say I don't know English so much :(

Answer 38

@TheViper Well you can your welcome instead of putting dude.

Answer 39

OK u r welcome :)

Answer 40

@TammyFlow :D

Answer 41

Ah okay, basically, you can do the drawing out version, but algebraically works just fine. A circle with a circle always ends up in a quadratic equation. Therefore compute the discriminant:

\[ \Large D= b^2-4ac\]
If D=0 you have found a tangent to the circle, it only meets the circle at exactly one point

If D>0 then you can compute the quadratic equation normally and you will obtain two solutions, therefore you get two points where your line meets the circle, it transverses it.

Finally if D<0 then there are only complex roots, which means that there are no solutions in \( \mathbb{R}\) and your line never meets the circle at any given point.

Answer 42

@TheViper ;). See that's better

Answer 43

So i guess ill try that way also. But for some problems ill just draw it out

Answer 44

whatever suits you the best (-:

Answer 45

Thanks for your help. I'll see which one is simple.

Answer 46

Or you can just use a graphing calculator and graph both equations. Graphing is the fastest way to see whether the line and the circle actually intersect. The reason why I didn't graph it first is because I expected the two equations to have a solution. When the solutions were complex, for some reason it didn't occur to me it was because the line and the circle did not intersect. Must've been smoking something yesterday.

Answer 47

No it was just late. Now i know graphing is easier then anything depending on the equation. But thanks for all your help.

Answer 48

@TammyFlow It's always useful to sketch. It may not be necessary, but it keeps your feet on the ground.

And, FWIW :) you're not a dude, your're a dudette. At least, that's what we say in my family..

Answer 49

@telliott99 Thanks Dude now i have earn a new nickname. And it is more useful to graph it because it save time depending on the equation.

Answer 50

Hope it's ok. Graphing is really the first thing to do almost always.

Answer 51

Ok. Thanks