C1 circles?

There is no k^4

You have turned a k into a k^2 in the x part

Reply 2

OP

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?

yes

OP

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 ??

Reply 5

x^2 + k^2x^2 - 4x - 4kx + 7 = (k^2+1)x^2 - 4(k+1)x + 7[br][br]b^2-4ac = 16(k+1)^2 - 28(k^2+1) = 16k^2 + 32k + 16 - 28k^2 - 28 [br][br]so 12k^2 - 32k +12 = 0[br][br]so 3k^2 - 8k + 3 = 0

Reply 6

[br]k = {(8\pm\sqrt{64-36})}/{6}[br][br]= [8\pm2\sqrt7]/6[br]

giving as req

(edited 12 years ago)

Reply 7

OP

Original post by TenOfThem

x^2 + k^2x^2 - 4x - 4kx + 7 = (k^2+1)x^2 - 4(k+1)x + 7[br][br]b^2-4ac = 16(k+1)^2 - 28(k^2+1) = 16k^2 + 32k + 16 - 28k^2 - 28 [br][br]so 12k^2 - 32k +12 = 0[br][br]so 3k^2 - 8k + 3 = 0

Ah, thanks so I think it's the 4(k+1)x, as I forgot the 4. :s-smilie:

Thanks! I'll see if I can do the second question now :smile:

Reply 8

OP

Original post by TenOfThem

[br]k = {(8\pm\sqrt{64-36})}/{6}[br][br]= [8\pm2\sqrt7]/6[br]

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?

Reply 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

\frac{4a+6b}{2} = \frac{4a}{2} + \frac{6b}{2} = 2a + 3b

Reply 10

OP

Original post by TenOfThem

no you divide the whole lot

\frac{4a+6b}{2} = \frac{4a}{2} + \frac{6b}{2} = 2a + 3b

Oh yeah sorry, just remembered that you only do that if it is x * y. not x+y. :colondollar:

Reply 11

OP

Original post by TenOfThem

no you divide the whole lot

\frac{4a+6b}{2} = \frac{4a}{2} + \frac{6b}{2} = 2a + 3b

Erm, could you help with this question as well? :s-smilie:

"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?)

Reply 12

Ok so completing the square does give

(x-2)^2 - 4 + (y-1)^2 - 1 - 5 = 0[br]

giving

(x-2)^2 + (y-1)^2 = 10[br][br]

centre = (2,1) and radius =

\sqrt{10}

(edited 12 years ago)

Reply 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?

Reply 14

OP

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? :/

Reply 15

yes the discriminant :smile:

The -5 that is already there is already there

In the equation you have three bits

[x^2-4x], [y^2-2y], and [-5]

x^2-4x = (x-2)^2-4[br]y^2-2y = (y-1)^2-1[br]-5 = -5[br]

Reply 16

OP