Question

How to get x = -3 and y = 5 using these equations?
3x-y+14=0
121+4xy-6x-28y^2+y^2+4x^2=0
-14x -72y+149-4xy+4y^2+x^2=0

Answer 1

Ok so the problem is the circle passes through 1 point (-1,6) and tangent to two lines (2x+y+6=0) (x-2y+8=0) and I got values 3x-y+14=0, 121+4xy-6x-28y^2+y^2+4x^2 and -14x -72y+149-4xy+4y^2+x^2

Answer 2

but I have no Idea how I'm supposed to work these out, I plugged it in a graphin thingy and I got the correct coordinates with them

Answer 3

I'm trying to find the center btw

Answer 4

the center I got is (-3,5)

Answer 5

Here's the graphical representation of the equations I got, I just need to know how to solve for x and y https://www.desmos.com/calculator/cl7wozqmvg

Answer 6

the equation of the tangents go through A and B - then you can deduce the equations of the radii passing through the center O since they are perpendicular to the tangents. Solving these 2 equations s will give you the coordinates of the center

Answer 7

oh - you also need to know the coordinates of A and B ....

Answer 8

I'm not sure if you have enough info to do this question

Answer 9

sorry i gotta go right now...

Answer 10

ah yeah, I got the equations actually
3x-y+14=0
121+4xy-6x-28y^2+y^2+4x^2=0
-14x -72y+149-4xy+4y^2+x^2=0
solving any 2 of these will get you x and y so says my professor, but I really don't know how.

Answer 11

This is what those three equations should look like, but I really don't know how to get (-3,5)

Answer 12

heyy still stuck on this ?

Answer 13

yes ;~;

Answer 14

Let the equation of required circle be \((x-h)^2 + (y-k)^2 = r^2\)
since this circle passes through \((-1, 6)\), we have :
\[(-1-h)^2 + (6-k)^2 = r^2 \tag{1}\]