# 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?

Sep 5, 2016

There are two solutions as explained below.

#### Explanation:

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

So

${C}_{1} \to {y}^{2} + {x}^{2} - 4 x - 6 y + 9 = 0$ for
${C}_{1} \to {\left(x - {x}_{1}\right)}^{2} + {\left(y - {y}_{1}\right)}^{2} = {r}_{1}^{2}$
${C}_{2} \to {y}^{2} + {x}^{2} + 10 x - 16 y + 85 = 0$ for
${C}_{2} \to {\left(x - {x}_{2}\right)}^{2} + {\left(y - {y}_{2}\right)}^{2} = {r}_{2}^{2}$
${L}_{1} \to x + 2 y + 15 = 0$
${L}_{1} \to p = {p}_{3} + {\lambda}_{3} {\vec{v}}_{3}$

After reduction, we have

${p}_{1} = \left(2 , 3\right) , {r}_{1} = 2$
${p}_{2} = \left(- 5 , 8\right) , {r}_{2} = 2$
${p}_{3} = \left(- 15 , 0\right) , {\vec{v}}_{3} = \left(1 , 2\right)$

Now, given

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

find

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

such that

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

Here $p = \left(x , y\right)$ and ${p}_{0} = \left({x}_{0} , {y}_{0}\right)$

We can stablish the following retationships

$\left\lVert {p}_{0} - {p}_{1} \right\rVert = {r}_{0} + {r}_{1}$
$\left\lVert {p}_{o} - {p}_{2} \right\rVert = {r}_{0} + {r}_{2}$

if ${p}_{t} \in {L}_{1}$ is a tangency point then

$\left\langle{p}_{t} - {p}_{0} , {\vec{v}}_{3}\right\rangle = 0$
$\left\lVert {p}_{t} - {p}_{0} \right\rVert = {r}_{0}$

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.