# Thread: Intersection Of A Line And Circle...

1. ## Intersection Of A Line And Circle...

Can anybody explain to me what to do?

x-2y=0 and (x+1)^2 + (y+3)^2 =5

My attempt:

sqrt5 = 2.24

find y:
x-2y=0
y=x/-2

substitute:
(x+1)^2 + (x/-2 + 3)^2 = 5
....to 1.25x^2-1x+5=0
... imaginary.

Eek. Smart people help?

2. Originally Posted by AlphaRock
Double post. Sorry. Computer been getting laggy.
I notice this after the fact that

I think you have an algeba error

$\displaystyle y=\frac{1}{2}x$

Please ignore the graph as I didn't notice the error until after I posted it.

Good luck.

3. Originally Posted by AlphaRock
x-2y=0
(x+1)^2+(y+3)^2=5

My attempt:

find y
y=x/-2

sqrt5 = 2.24

(x+1)^2 + ((x/-2)+3)^2=5
...1.25x^2-1x+10-5=5
...imaginary?wtf
Easier to substitute x = 2y into the circle: y^2 + 2y + 1 = 0. One real solution, blue sky.

4. Originally Posted by TheEmptySet
here is a graph to explain what you have found.

There is not intersection.

When you get not real solutions they curves don't intersect
Ah thank you!!!

I only have one more question for you:

For finding the distance from the point A(5,7) to the line joining B(-2,1) and C(4,-3).

Is it 8.94?

I did this by 1. Finding the midpoint of BC, then using the distance formula from (1,-1)(midpoint) to A(5,7) which equals 8.94

Is there an easier way to do this?
Because hypothetically speaking, if it were A(-4,0), and B(2,0), and C(-4,1), I cannot get the midpoint, then use the distance formula to get it correct. It would be wrong.

5. Originally Posted by TheEmptySet
I notice this after the fact that

I think you have an algeba error

$\displaystyle y=\frac{1}{2}x$

Please ignore the graph as I didn't notice the error until after I posted it.

Good luck.
Thanks. Nice catch.

When I graph it, there's still no intersection.

But algebraically i got, (-2,5) (-2,-1)... could somebody clear the fog for this question and my other question?

6. You can't assume the midpoint of those two points will be the shortest distance from that point to it. You could probably draw a situation in which you'll see this isn't the case. Also, it's not a line segment that you're dealing with but the entire line.

If you recall the perpendicular distance formula from a point to a line, this represents the shortest distance: $\displaystyle D = \frac{Ax_{0} + By_{0} + C}{\sqrt{A^{2} + B^{2}}}$
where $\displaystyle (x_{0},y_{0})$ is your point and $\displaystyle Ax + By + C = 0$ is the standard form of your line.

So, you find the equation of the line by finding the slope and plugging it into the point-slope equation of a line: $\displaystyle y - y_{0} = m(x-x_{0})$ and put it in the form $\displaystyle Ax + By + C = 0$ with the help of either points B or C.

Then plug it into your distance formula.

Now, I didn't plot the points out or anything so there may be a more elegant way of doing this but this works just as well.

7. Originally Posted by o_O
You can't assume the midpoint of those two points will be the shortest distance from that point to it. You could probably draw a situation in which you'll see this isn't the case. Also, it's not a line segment that you're dealing with but the entire line.

If you recall the perpendicular distance formula from a point to a line, this represents the shortest distance: $\displaystyle D = \frac{Ax_{0} + By_{0} + C}{\sqrt{A^{2} + B^{2}}}$
where $\displaystyle (x_{0},y_{0})$ is your point and $\displaystyle Ax + By + C = 0$ is the standard form of your line.

So, you find the equation of the line by finding the slope and plugging it into the point-slope equation of a line: $\displaystyle y - y_{0} = m(x-x_{0})$ and put it in the form $\displaystyle Ax + By + C = 0$ with the help of either points B or C.

Then plug it into your distance formula.

Now, I didn't plot the points out or anything so there may be a more elegant way of doing this but this works just as well.
Thanks!

Did you get 8.88?

And how do you graph this...?

There's another way to find the shortest distance by:
1. Get two perpendicular lines
2. Find out the intersects of both
3. Use distance formulas of intersections

Now. I find your way is more easier to find the distance. But how do you graph it? Last test I did, I had to stick a random line on the page.

Like:
I = line
_________
I
I_________

8. Originally Posted by AlphaRock
Thanks. Nice catch.

When I graph it, there's still no intersection.

But algebraically i got, (-2,5) (-2,-1)... could somebody clear the fog for this question and my other question?
(-2,-1) is the only solution. The line touches the circle... its a tangent.How did you get (-2,5)?

For finding the distance from the point A(5,7) to the line joining B(-2,1) and C(4,-3).
o_O's method is the easiest way to do it. And it is derived using your idea:
There's another way to find the shortest distance by:
1. Get two perpendicular lines
2. Find out the intersects of both
3. Use distance formulas of intersections

9. Originally Posted by Isomorphism
(-2,-1) is the only solution. The line touches the circle... its a tangent.How did you get (-2,5)?

o_O's method is the easiest way to do it. And it is derived using your idea:
Thanks, you're right. It's only (-2, -1).

Good eye.

10. Hi im new. But i thought i better point out that you are makeing a mistake, because that graph is correct the line and the circle dont cross so there is no intersept.

11. Originally Posted by AlphaRock
Can anybody explain to me what to do?

x-2y=0 and (x+1)^2 + (y+3)^2 =5

My attempt:

sqrt5 = 2.24

find y:
x-2y=0
y=x/-2

substitute:
(x+1)^2 + (x/-2 + 3)^2 = 5
....to 1.25x^2-1x+5=0
... imaginary.

Eek. Smart people help?
I guess my reply at: http://www.mathhelpforum.com/math-he...ne-circle.html

is out in the cold

12. Originally Posted by N.A.G
Hi im new. But i thought i better point out that you are makeing a mistake, because that graph is correct the line and the circle dont cross so there is no intersept.
Sorry but you're the one making the mistake. The line and circle do intersect, at the point (-2, -1).

As TheEmptySet has already stated, the graphs are not correct. In particular, the line cannot be correct - it's gradient is negative and the line x = 2y has a positive gradient.