Question

Can you help me solve this system: 15k^2 + 24k + 8kn + n^2 - 6n + 8 = 0
8k^2 - 3k - 4kn - n^2 - 2n + 12 = 0

Answer 1

\[15k^{2} + 24k + 8kn + n^{2} - 6n + 8 = 0\]
\[8k^{2} - 3k - 4kn - n^{2} - 2n + 12 = 0\]

Answer 2

http://www.wolframalpha.com/input/?i=solve%28+15+k^2+%2B+24+k+%2B+8+k+n+%2B+n^2+-+6+n+%2B+8++%3D+0+and+++8+k^2+-+3+k+-+4+k+n+-+n^2+-+2+n+%2B+12+%3D+0%29

Answer 3

Ok, but how do you solve it algebraically?

Answer 4

Are you sure that your equations are correct? I do not believe there is an easy way to do it by hand

Answer 5

Well, even if the coefficients aren't correct (though I believe they are), do you know the algorithm for solving these kinds of equations?

Answer 6

Or could you help me with the original problem: Finding the common tangent lines to circles
\[(x-4)^{2} + (y - 3)^{2} = 1\]
\[(x-2)^{2} + (y + 1)^{2} = 9\]

Answer 7

I substituted y = kx + n to both of theses equations and got that system with k and n. Is there some other way to find those tangent lines?

Answer 8

*these

Answer 9

http://math.stackexchange.com/questions/211538/common-tangent-to-two-circles

Answer 10

http://mathworld.wolfram.com/Circle-CircleTangents.html

Answer 11

That's great thank you :)

Answer 12

yw