1. Maths Help

I am really stuck on this question...

I have my Maths Prelim in a couple of days and I think I am ok with most things however I am stumped by this one question.

The line y-2x=k is tangent to the circle x^2+y^2-2x-4=0, where k>0.

Find the value of k.

I have got to x^2+(k-2x)^2-2x-4=0, I know that I need to apply that to the discriminant (b^2-4ac) but I don't know how to get that into the discriminant.

Thanks For Any help,

Bumfluff

2. y - 2x = k
y - 2x + 2x = k + 2x <== Note the '+' in place of the '-' you have.
y = k + 2x

Maybe this will help some.

3. Sorry I have written that wrong, it is y+2k, but that still doesnt help me.

4. This is cracked out, probably not the right way to go about doing things. You can complete the stuff to find k.

Code:
```slope = 2

d/dx [x^2 + y^2 - 2x - 4] = 2x + 2y dy/dx - 2

(1 - x) / y = dy/dx = 2

(1 - x) / 2 = y = k + 2x

k = (1 - 5x) / 2

x^2 + (k+2x)^2 - 2x - 4 = 0

x^2 - 2x + 1 + (k+2x)^2 - 4 = 1

(x - 1)^2 + (k + 2x)^2 = 5

k = sqrt( 5 - (x - 1)^2 ) - 2x

(1 - 5x) / 2 = sqrt( 5 - (x - 1)^2 ) - 2x

(1 - x)^2 / 4 = 5 - (x - 1)^2

(1 - x)^2 = 4

1 - x = +- 2

x = 3, -1```

5. I like Tonto's way There isn't half enough calculus used these days!

Here's a nicer looking way (nicer cause it's all math-typed):

Attachment 7136

6. twomers, the normal line should intersect the circle at two points, right?
5y^2-5=0 <=> y = +-1

7. Yes. You're right of course. I should have specified, but just took the answer that gave the right answer, if you get me.

8. I'm not sure I do get you. There are two possible values of k, if I haven't misunderstood anything.

9. Let's see, I need to make some more substance to make this post 4 or more characters so that is what I am doing. K is greater than or equal to zero.

10. Well the advantage twomers had was that he knew the answer to be 7 from talking on MSN so he was working towards a known solution. But I got my maths teacher to go thorugh it today. And he showed me using the discriminant.

11. There's a much simpler solution that involves solving only one quadratic equation.. to obtain points where the line crosses the circle, substitute y = 2x + k in the circle's equation x&#178;+y&#178;-2x+4 = 0. You get

x&#178;+(2x+k)&#178;-2x+4 = 0, which is simplified to
5x&#178; + (4k-2)x + k&#178;-4 = 0.

Now we are interested in the values of k for which the above equation quadratic in x has one and only one solution, because only the tangent lines intersect the circle in only one point--other lines either don't interect at all, or intersect in two different points. A quadratic equation has one and only one solution if and only if its determinant is 0, so we get

(4k-2)&#178; - 4 * 5 * (k&#178;-4) = 0. Dividing by 4
(2k-1)&#178; - 5 * (k&#178;-4) = 0, which simplifies to
(4k&#178;-4k+1) - 5k&#178; + 20 = 0, or
-k&#178; - 4k + 21 = 0.

Now this quadratic equation can be solved by any method you like, and get the solutions k = -7 or k = 3.