Let the radius of circle B be MB=R2, because it is circumscribed, so MA=R 1+R2.

Therefore, MA-MB=R 1=2.

Is the right branch of hyperbola, and the focus is (-√ 2,0) (√ 2,0).

2a=2，a = 1；

c =√2；

b=√(c^2-a^2)= 1

So the analytical formula is:

x？ -Really? = 1(x >0), which is the right branch of hyperbola.

2. simultaneous equations of hyperbola and straight line l,

Because the straight line and curve c have two different intersections,

That is, there are two different intersections between the straight line and the right branch of the hyperbola.

x？ -Really? = 1

Y=kx+ 1 simultaneously.

Get ( 1-k？ )x？ -2kx-2=0

There are two different intersections with the right branch, that is, both roots are greater than 0,

that is

x 1+x2=2k/( 1-k？ )& gt0

x 1*x2= - 2/( 1-k？ )& gt0

δ=(2k)？ +8( 1-k？ )& gt0

Solve it separately,

(-infinity,-1)(0, 1)

(-infinity,-1), (1, positive infinity)

(-√2,√2)

Take the intersection of the three and get -√ 2.

I hope it helps you ~