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#?

Question

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#?

Claire Akin · Accepted Answer

See below.

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

#L_1->x+3y=0# to #L-1->p =p_1+lambda_1 vec v_1 #
#L_2->3x + y +8 =0# to #L_2->p=p_2+lambda_2 vec v_2#
#C_1-> x^2+y^2-10x-6y+30=0# to #C_1->norm(p-p_3)=r_3#

Here
#p_1 = (0, 0)#
#p_2 = (-2, -1)#
#vec v_1 = (1, 3)#
#vec v_2 = (3, 1)#
#p_3 = (5, 3)#
#r_3 = 2#

Now, given

#L_1->p=p_1+lambda_1 vec v_1#
#L_2->p=p_2+lambda_2 vec v_2#
#C_1->norm(p-p_3)=r_3#

and

#C->norm(p-p_0) = r#

with #p=(x,y)#

If #C# is tangent to #L_1, L_2# and #C_1# then

#p_0 in L_{12}# where

#L_{12}->p_(12)+lambda_(12) vec v_(12)#

with #p_(12) = L_1 nn L_2# and #vec v_(12) = vec v_1/norm(vec v_1) + vec v_2/norm(vec v_2)#

Other conditions for #p_0# and #r# are

#norm(p_(12)-p_0)^2- << p_(12)-p_0, vec v_1/norm(vec v_1) >>^2 = r^2#
#norm(p_0-p_3) = r + r_3#

but

#p_0=p_(12)+lambda_(12)vec v_(12)#

so

#lambda_(12)^2norm(vec v_(12))^2-lambda_(12)^2 << vec v_(12),vec v_1/norm(vec v_1) >> ^2= r^2#

and

#norm(p_(12)+lambda_(12) vec v_(12) - p_3)^2=(r+r_3)^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#

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.

Carson Arnold · Answer

undefined To find the circle tangent to the lines $ L_1: x + 3y = 0 $, $ L_2: 3x + y + 8 = 0 $, and the circle $ C_1: x^2 + y^2 - 10x - 6y + 30 = 0 $, we need to follow these steps:

1. Find the point of intersection of $ L_1 $ and $ L_2 $ to determine the center of the circle.
2. Calculate the distance between the center of the circle and the given circle $ C_1 $ to determine the radius of the circle.

First, let's find the point of intersection of $ L_1 $ and $ L_2 $:

Solving the system of equations formed by $ L_1 $ and $ L_2 $, we get the point of intersection $ (x_0, y_0) $.

Next, calculate the distance between $ (x_0, y_0) $ and the center of the circle $ C_1 $ using the formula for the distance between two points:

$$ d = \sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2} $$

where $ (x_1, y_1) $ is the center of $ C_1 $.

This distance represents the radius of the circle $ C $. Therefore, the equation of the circle tangent to $ L_1 $, $ L_2 $, and $ C_1 $ is:

$$ (x - x_0)^2 + (y - y_0)^2 - r^2 = 0 $$

where $ (x_0, y_0) $ is the center of the circle and $ r $ is the radius of the circle.

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

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#?