# Given L_1->x+3y=0, L_2=3x+y+8=0 and C_1=x^2+y^2-10x-6y+30=0, determine C->(x-x_0)^2+(y-y_0)^2-r^2=0 tangent to L_1,L_2 and C_1?

Sep 6, 2016

See below.

#### Explanation:

Firstly we will pass the geometrical objects to a more convenient representation.

${L}_{1} \to x + 3 y = 0$ to $L - 1 \to p = {p}_{1} + {\lambda}_{1} {\vec{v}}_{1}$
${L}_{2} \to 3 x + y + 8 = 0$ to ${L}_{2} \to p = {p}_{2} + {\lambda}_{2} {\vec{v}}_{2}$
${C}_{1} \to {x}^{2} + {y}^{2} - 10 x - 6 y + 30 = 0$ to ${C}_{1} \to \left\lVert p - {p}_{3} \right\rVert = {r}_{3}$

Here
${p}_{1} = \left(0 , 0\right)$
${p}_{2} = \left(- 2 , - 1\right)$
${\vec{v}}_{1} = \left(1 , 3\right)$
${\vec{v}}_{2} = \left(3 , 1\right)$
${p}_{3} = \left(5 , 3\right)$
${r}_{3} = 2$

Now, given

${L}_{1} \to p = {p}_{1} + {\lambda}_{1} {\vec{v}}_{1}$
${L}_{2} \to p = {p}_{2} + {\lambda}_{2} {\vec{v}}_{2}$
${C}_{1} \to \left\lVert p - {p}_{3} \right\rVert = {r}_{3}$

and

$C \to \left\lVert p - {p}_{0} \right\rVert = r$

with $p = \left(x , y\right)$

If $C$ is tangent to ${L}_{1} , {L}_{2}$ and ${C}_{1}$ then

${p}_{0} \in {L}_{12}$ where

${L}_{12} \to {p}_{12} + {\lambda}_{12} {\vec{v}}_{12}$

with ${p}_{12} = {L}_{1} \cap {L}_{2}$ and ${\vec{v}}_{12} = {\vec{v}}_{1} / \left\lVert {\vec{v}}_{1} \right\rVert + {\vec{v}}_{2} / \left\lVert {\vec{v}}_{2} \right\rVert$

Other conditions for ${p}_{0}$ and $r$ are

${\left\lVert {p}_{12} - {p}_{0} \right\rVert}^{2} - {\left\langle{p}_{12} - {p}_{0} , {\vec{v}}_{1} / \left\lVert {\vec{v}}_{1} \right\rVert\right\rangle}^{2} = {r}^{2}$
$\left\lVert {p}_{0} - {p}_{3} \right\rVert = r + {r}_{3}$

but

${p}_{0} = {p}_{12} + {\lambda}_{12} {\vec{v}}_{12}$

so

${\lambda}_{12}^{2} {\left\lVert {\vec{v}}_{12} \right\rVert}^{2} - {\lambda}_{12}^{2} {\left\langle{\vec{v}}_{12} , {\vec{v}}_{1} / \left\lVert {\vec{v}}_{1} \right\rVert\right\rangle}^{2} = {r}^{2}$

and

${\left\lVert {p}_{12} + {\lambda}_{12} {\vec{v}}_{12} - {p}_{3} \right\rVert}^{2} = {\left(r + {r}_{3}\right)}^{2}$
${\left\lVert {p}_{12} - {p}_{3} \right\rVert}^{2} + 2 {\lambda}_{12} \left\langle{p}_{12} - {p}_{3} , {\vec{v}}_{12}\right\rangle + {\lambda}_{12}^{2} {\left\lVert {\vec{v}}_{12} \right\rVert}^{2} = {\left(r + {r}_{3}\right)}^{2}$

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

$\left(\begin{matrix}{\lambda}_{12} = 1.64171 & r = 1.31337 \\ {\lambda}_{12} = 8.00809 & r = 6.40647\end{matrix}\right)$

The attached plot shows the answer in red and the initial elements in black.