Firstly we will pass the geometrical objects to a more convenient representation.
L_1->x+3y=0L1→x+3y=0 to L-1->p =p_1+lambda_1 vec v_1 L−1→p=p1+λ1→v1
L_2->3x + y +8 =0L2→3x+y+8=0 to L_2->p=p_2+lambda_2 vec v_2L2→p=p2+λ2→v2
C_1-> x^2+y^2-10x-6y+30=0C1→x2+y2−10x−6y+30=0 to C_1->norm(p-p_3)=r_3C1→∥p−p3∥=r3
Here
p_1 = (0, 0)p1=(0,0)
p_2 = (-2, -1)p2=(−2,−1)
vec v_1 = (1, 3)→v1=(1,3)
vec v_2 = (3, 1)→v2=(3,1)
p_3 = (5, 3)p3=(5,3)
r_3 = 2r3=2
Now, given
L_1->p=p_1+lambda_1 vec v_1L1→p=p1+λ1→v1
L_2->p=p_2+lambda_2 vec v_2L2→p=p2+λ2→v2
C_1->norm(p-p_3)=r_3C1→∥p−p3∥=r3
and
C->norm(p-p_0) = rC→∥p−p0∥=r
with p=(x,y)p=(x,y)
If CC is tangent to L_1, L_2L1,L2 and C_1C1 then
p_0 in L_{12}p0∈L12 where
L_{12}->p_(12)+lambda_(12) vec v_(12)L12→p12+λ12→v12
with p_(12) = L_1 nn L_2p12=L1∩L2 and vec v_(12) = vec v_1/norm(vec v_1) + vec v_2/norm(vec v_2)→v12=→v1∥∥→v1∥∥+→v2∥∥→v2∥∥
Other conditions for p_0p0 and rr are
norm(p_(12)-p_0)^2- << p_(12)-p_0, vec v_1/norm(vec v_1) >>^2 = r^2∥p12−p0∥2−⟨p12−p0,→v1∥∥→v1∥∥⟩2=r2
norm(p_0-p_3) = r + r_3∥p0−p3∥=r+r3
but
p_0=p_(12)+lambda_(12)vec v_(12)p0=p12+λ12→v12
so
lambda_(12)^2norm(vec v_(12))^2-lambda_(12)^2 << vec v_(12),vec v_1/norm(vec v_1) >> ^2= r^2λ212∥∥→v12∥∥2−λ212⟨→v12,→v1∥∥→v1∥∥⟩2=r2
and
norm(p_(12)+lambda_(12) vec v_(12) - p_3)^2=(r+r_3)^2∥∥p12+λ12→v12−p3∥∥2=(r+r3)2
norm(p_(12)-p_3)^2+2lambda_(12) << p_(12)-p_3, vec v_(12) >> +lambda_(12)^2 norm(vec v_(12))^2=(r+r_3)^2∥p12−p3∥2+2λ12⟨p12−p3,→v12⟩+λ212∥∥→v12∥∥2=(r+r3)2
The essential set of equations to obtain the solution is
{(lambda_(12)^2(norm(vec v_(12))^2 - << vec v_(12),vec v_1/norm(vec v_1) >> ^2)= r^2),
(norm(p_(12)-p_3)^2+2lambda_(12) << p_(12)-p_3, vec v_(12) >> +lambda_(12)^2 norm(vec v_(12))^2=(r+r_3)^2):}
two equations and two incognitas r, lambda_(12)
Solving for r, lambda_(12) we obtain
((lambda_(12) = 1.64171, r = 1.31337),(lambda_(12) = 8.00809, r = 6.40647))
The attached plot shows the answer in red and the initial elements in black.
![enter image source here]()
