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#?

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.