# Analytic Geometry

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

• Analytic Geometry -

slope of x-3y - 11= 0 is 1/3
slope of 3x - y - 9 = 0 is 3

A little known theorem states that if two lines have perpendicular slopes, like our case, then the angle bisector of their obtuse angle between them has a slope of -1

intersection of the two lines
x-3y = 11 and
3x - y = 9

2nd times 3:
9x - 3y = 27
x - 3y = 11, the first equation
-----------
8x = 16
x = 2
sub into 1st:
2 - 3y = 11
-3y = 9
y = -3
So they intersect at (2, -3)

the centre of the circle must lie on this angle bisector , having slope -1 and passing through (2, -3)
equation of that angle bisector:
y = -x + b , with (2,3) on it, so
-3 = -2 + b ---> b = -1
for y = -x - 1

the centre must be the intersection of that line and x+2y = -19
x + 2(-x-1) = -19
x - 2x - 2 = -19
-x = -17
x = 17
then y = -17 - 1 = -19

the centre is (17 , -19)
distance to one of the lines would be the radius.
distance to x - 3y - 11 = 0 is
|17 -+ 3(-19) - 11|/√(1^2 + (-3)^2) = 60/√10

FINALLY:
equation of circle
(x-17)^2 + (y+19)^2 = (60/√10)^2

(x-17)^2 + (y+19)^2 = 360

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