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?

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- May 5th 2008, 08:40 PMAlphaRockIntersection 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? - May 5th 2008, 09:18 PMTheEmptySet
- May 5th 2008, 09:57 PMmr fantastic
- May 5th 2008, 10:27 PMAlphaRock
Ah thank you!!! (Hi)

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. - May 5th 2008, 10:39 PMAlphaRock
- May 5th 2008, 10:40 PMo_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. - May 5th 2008, 10:54 PMAlphaRock
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_________ - May 5th 2008, 11:35 PMIsomorphism
(-2,-1) is the only solution. The line touches the circle... its a tangent.How did you get (-2,5)?

Quote:

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:

Quote:

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

- May 5th 2008, 11:51 PMAlphaRock
- May 6th 2008, 12:13 AMN.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.

- May 6th 2008, 02:42 AMmr fantastic
I guess my reply at: http://www.mathhelpforum.com/math-he...ne-circle.html

is out in the cold (Speechless) - May 6th 2008, 03:11 AMmr fantastic
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.