What is the equation of a circle touching the lines x-3y-11=0 and 3x-y-9=0 and having its center on the line x+2y+19=0?

Let O(x, y) is the center of the circle that lies on the bisector of the angle between the lines x-3y-11=0 and 3x-y-9=0 then

|x - 3y - 11|/(1^2 + (-3)^2)^0.5 = |3x - y - 9|/(3^2 + (-1)^2)^0.5

|x - 3y - 11|/10^0.5 = |3x - y - 9|/10^0.5

x - 3y - 11 = 3x - y - 9

2x + 2y + 2 = 0

x + y + 1 = 0 is one bisector of the angle between the lines x-3y-11=0 and 3x-y-9=0

x - 3y - 11 = - 3x + y + 9

4x - 4y - 20 = 0

x - y - 5 = 0 is the other bisector of the angle between the lines x-3y-11=0 and 3x-y-9=0

Since the center lies on x+2y+19=0 we find

x+2y+19=0

x + y + 1 = 0

x = 17

y = -18

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and

x+2y+19=0

x - y - 5 = 0

x = - 3

y = - 8

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we found the center of the circle O1(17, -18) and O2(-3, -8)

lets find the radius of the circle

for O1(17, -18) the distance from O1 to x-3y-11=0 is

r = |17 -3*(-18) - 11|/(1^2 + (-3)^2)^0.5 = 60/(10)^0.5 =

r^2 = 360

the equation of the circle is

(x - 17)^2 + (y + 18)^2 = 360

for O2(-3, -8) the distance from O2 to x-3y-11=0 is

r = |(-3) -3*(-8) - 11|/(1^2 + (-3)^2)^0.5 = (10)^0.5

r^2 = 10

the equation of the circle is

(x + 3)^2 + (y + 8)^2 = 10

There are 2 circles that are touching the lines x-3y-11=0 and 3x-y-9=0 and having its center on the line x+2y+19=0

(x - 17)^2 + (y + 18)^2 = 360

(x + 3)^2 + (y + 8)^2 = 10

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