Given $C_{1} \to y^{2} + x^{2} - 4 x - 6 y + 9 = 0$ $C_1->y^2+x^2-4x-6y+9=0$ , $C_{2} \to y^{2} + x^{2} + 10 x - 16 y + 85 = 0$ $C_2->y^2+x^2+10x-16y+85=0$ and $L_{1} \to x + 2 y + 15 = 0$ $L_1->x+2y+15=0$ , determine $C \to {(x - x_{0})}_{0}^{2} + {(y - y_{0})}_{0}^{2} - r^{2} = 0$ $C->(x-x_0)^2+(y-y_0)^2-r^2=0$ tangent to $C_{1}, C_{2}$ $C_1,C_2$ and $L_{1}$ $L_1$ ?

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Given $C_{1} \to y^{2} + x^{2} - 4 x - 6 y + 9 = 0$ $C_1->y^2+x^2-4x-6y+9=0$ , $C_{2} \to y^{2} + x^{2} + 10 x - 16 y + 85 = 0$ $C_2->y^2+x^2+10x-16y+85=0$ and $L_{1} \to x + 2 y + 15 = 0$ $L_1->x+2y+15=0$ , determine $C \to {(x - x_{0})}_{0}^{2} + {(y - y_{0})}_{0}^{2} - r^{2} = 0$ $C->(x-x_0)^2+(y-y_0)^2-r^2=0$ tangent to $C_{1}, C_{2}$ $C_1,C_2$ and $L_{1}$ $L_1$ ?

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Answer 1 · 2016-09-05T10:21:11

There are two solutions as explained below.

Explanation:

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

So

$C_{1} \to y^{2} + x^{2} - 4 x - 6 y + 9 = 0$ $C_1->y^2+x^2-4x-6y+9=0$ for
$C_{1} \to {(x - x_{1})}_{1}^{2} + {(y - y_{1})}_{1}^{2} = r_{1}^{2}$ $C_1->(x-x_1)^2+(y-y_1)^2=r_1^2$
$C_{2} \to y^{2} + x^{2} + 10 x - 16 y + 85 = 0$ $C_2->y^2+x^2+10x-16y+85=0$ for
$C_{2} \to {(x - x_{2})}_{2}^{2} + {(y - y_{2})}_{2}^{2} = r_{2}^{2}$ $C_2->(x-x_2)^2+(y-y_2)^2=r_2^2$
$L_{1} \to x + 2 y + 15 = 0$ $L_1->x+2y+15=0$
$L_{1} \to p = p_{3} + λ_{3} {\to v}_{3}$ $L_1->p = p_3+lambda_3 vec v_3$

After reduction, we have

$p_{1} = (2, 3), r_{1} = 2$ $p_1=(2,3), r_1=2$
$p_{2} = (- 5, 8), r_{2} = 2$ $p_2=(-5,8), r_2=2$
$p_{3} = (- 15, 0), {\to v}_{3} = (1, 2)$ $p_3=(-15,0), vec v_3 =(1,2)$

Now, given

$C_{1} \to ∥ p - p_{1} ∥ = r_{1}$ $C_1->norm(p-p_1) = r_1$
$C_{2} \to ∥ p - p_{2} ∥ = r_{2}$ $C_2->norm(p-p_2)=r_2$ and
$L_{1} \to p_{3} + λ {\to v}_{3}$ $L_1->p_3+lambda vec v_3$

find

$C \to ∥ p - p_{0} ∥ = r_{0}$ $C->norm(p-p_0) = r_0$

such that

$C$ $C$ is tangent to $C_{1}, C_{2}$ $C_1,C_2$ and $L_{1}$ $L_1$

Here $p = (x, y)$ $p = (x,y)$ and $p_{0} = (x_{0}, y_{0})$ $p_0=(x_0,y_0)$

We can stablish the following retationships

$∥ p_{0} - p_{1} ∥ = r_{0} + r_{1}$ $norm(p_0-p_1)=r_0+r_1$
$∥ p_{o} - p_{2} ∥ = r_{0} + r_{2}$ $norm(p_o-p_2)=r_0+r_2$

if $p_{t} \in L_{1}$ $p_t in L_1$ is a tangency point then

$⟨ p_{t} - p_{0}, {\to v}_{3} ⟩ = 0$ $<< p_t-p_0, vec v_3 >> = 0$
$∥ p_{t} - p_{0} ∥ = r_{0}$ $norm(p_t-p_0)= r_0$

where

$p_{t} = p_{3} + λ {\to v}_{3}$ $p_t = p_3+lambda vec v_3$

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.

enter image source here

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