Question

Given `C_1->y^2+x^2-4x-6y+9=0`, `C_2->y^2+x^2+10x-16y+85=0` and `L_1->x+2y+15=0`, determine `C->(x-x_0)^2+(y-y_0)^2-r^2=0` tangent to `C_1,C_2` and `L_1`?

Answer 1

There are two solutions as explained below.

Explanation:

Firstly we will represent the geometric objects in a more convenient formulation.

So

`C_1->y^2+x^2-4x-6y+9=0` for
`C_1->(x-x_1)^2+(y-y_1)^2=r_1^2`
`C_2->y^2+x^2+10x-16y+85=0` for
`C_2->(x-x_2)^2+(y-y_2)^2=r_2^2`
`L_1->x+2y+15=0`
`L_1->p = p_3+lambda_3 vec v_3`

After reduction, we have

`p_1=(2,3), r_1=2`
`p_2=(-5,8), r_2=2`
`p_3=(-15,0), vec v_3 =(1,2)`

Now, given

`C_1->norm(p-p_1) = r_1`
`C_2->norm(p-p_2)=r_2` and
`L_1->p_3+lambda vec v_3`

find

`C->norm(p-p_0) = r_0`

such that

`C` is tangent to `C_1,C_2`and `L_1`

Here `p = (x,y)` and `p_0=(x_0,y_0)`

We can stablish the following retationships

`norm(p_0-p_1)=r_0+r_1`
`norm(p_o-p_2)=r_0+r_2`

if `p_t in L_1` is a tangency point then

`<< p_t-p_0, vec v_3 >> = 0`
`norm(p_t-p_0)= r_0`

where

`p_t = p_3+lambda vec v_3`

Finally we get the following equations

#{ (<< p_3+lambda vec v_3-p_0,vec v_3 >> = 0), (norm(p_3+lambda vec v_3-p_0)=r_0), (norm(p_0-p_1)=r_0+r_1), (norm(p_0-p_2)=r_0+r_2) :}#

so we have four equations and four incognitas `x_0,y_0,lambda,r_0`

Solving we obtain

#((r_0 = 8.17665, x_0 = -6.86077, y_0 = -2.00508, lambda = 0.825813), (r_0 = 11.6327, x_0 = 6.01912, y_0 = 16.0268, lambda = 10.6145))#

Attached the figure with the solutions in red and the initial geometric elements in black.