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# C1 circles? watch

1. Hi, I'm stuck on this question as I'm just unsure of how to progress, I'll show my working out and hopefully one of you nice people (or a few! ) can help

"The straight line with equation y=kx is a tangent to the circle x^2 + y^2 - 4x - 4y + 7 = 0. Find the possible values of k, giving your answers in the form a+b7^1/2."

So I substituted y=kx into the circle formula and got:

x^2 + (kx)^2 - 4x - 4(kx) + 7 =0

So: x^2 +k^2x^2 - 4x - 4kx + 7 = 0

x^2 (1+k^2) - 4x (1+k^2) + 7 = 0

And then I'm not sure what to do. I was going to use the discriminant (as it's a tangent b^2-4ac will = 0, but then I'd get a k^2*k^2 getting k^4? )

This is very similar to another question "The straight line with equation y=-3x+c is a tangent to the circle x^2+y^2-4x-2y-5=0. Find the possible values of c", but hopefully after understanding 1. I shall be able to do 2 as well Thanks!
2. There is no k^4

You have turned a k into a k^2 in the x part
3. (Original post by TenOfThem)
There is no k^4

You have turned a k into a k^2 in the x part
:O oh yeah! So am I doing it right?
4. yes
5. (Original post by TenOfThem)
yes
Erm still, it's wrong. As Then I have:

(1-k)^2 - 4(1+k^2)(7) = 0

So 1 + k^2 - 2k - 28 + 28k^2 = 0

29k^2 - 2k -27 =0

Quadratic formula...I then get 2+/-6 root87 (over) 58.

But it's meant to be 4/3 +/- 1/3 root7 ??

6. giving as req
7. (Original post by TenOfThem)
Ah, thanks so I think it's the 4(k+1)x, as I forgot the 4. Thanks! I'll see if I can do the second question now
8. (Original post by TenOfThem)

giving as req
If that was the answer, then couldn't you just divide the 8 by 6 to get 4/3, but then wouldn't it be 4/3 +/- 2 root 7 but the answer is 4/3 =/- 1/3 root 7 (which makes sense if you divide the whole numerator by 6, but don't you just divide one part of it? i.e. either the 4 or the 2root7?
9. (Original post by MedicalMayhem)
If that was the answer, then couldn't you just divide the 8 by 6 to get 4/3, but then wouldn't it be 4/3 +/- 2 root 7 but the answer is 4/3 =/- 1/3 root 7 (which makes sense if you divide the whole numerator by 6, but don't you just divide one part of it? i.e. either the 4 or the 2root7?
no you divide the whole lot

10. (Original post by TenOfThem)
no you divide the whole lot

Oh yeah sorry, just remembered that you only do that if it is x * y. not x+y.
11. (Original post by TenOfThem)
no you divide the whole lot

Erm, could you help with this question as well?

"The straight line with equation y=-3x+c is a tangent to the circle x^2+y^2-4x-2y-5=0. Find the possible values of c"

Well I'm just unsure how you would work out the c. As for co-ordinates where the lines intercept you could substiute one into another, but for this? I thought you might find the possible y and x values and then you get c, but I'm not too sure how to get the points? :/

Would you get the centre of the circle and the radius. So:

(x-2)^2 + (y-1)^2 -5 (and do you also do -4 - 1 as well, from expanding the brackets? As I remember (i think) that you don't but never learnt why...). So is it (x-2)^2 + (y-1)^2 -5 -4 -1 (or is it without the -4, -1)?

And then you get the center as 2,1, but then the radius is 0? :/ (as -4,-1 (from the conpleting the square bit) + 5 = 0?)
12. Ok so completing the square does give

giving

centre = (2,1) and radius =
13. But that is not how you do this problem

Substitute y = -3x+c into the original equation

You will have a quadratic in x that will give the values where the line and the circle meet

Since the line is a tangent how many solutions will this equation have?
14. (Original post by TenOfThem)
But that is not how you do this problem

Substitute y = -3x+c into the original equation

You will have a quadratic in x that will give the values where the line and the circle meet

Since the line is a tangent how many solutions will this equation have?
1 (the discriminant bit?).

But with your previous post, isn't the -5 which is already there shows the (-4,-1) from the completing the square? So would it be radius as root5? :/
15. yes the discriminant

The -5 that is already there is already there

In the equation you have three bits

16. (Original post by TenOfThem)
yes the discriminant

The -5 that is already there is already there

In the equation you have three bits

Ah okay, that's what I originally thought, but my other friend confused me. So for that question, I assume you just find the points where the line and curve cross, then substitute into the line (y=-3x+c) to get c. So seems easy enough, but then how would you go about creating a simultaneous equation with a c? :/

EDIT, or could you just explain what would be the first step to work it out?

17. A tangent so you only want one solution
18. (Original post by TenOfThem)

A tangent so you only want one solution
Oh so you actually can insert c But why does it say possible 'values' when you say you only want one solution?
19. One solution for x because there is only one point where the tangent and the circle meet

This will give 2 values of c

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