Posted by John on .
What is the equation of the circle touching the lines x3y11=0 and 3xy9=0 having its center on the line x+2y+19=0 .

Analytic Geometry 
Reiny,
slope of x3y  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
x3y = 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(x1) = 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
(x17)^2 + (y+19)^2 = (60/√10)^2
(x17)^2 + (y+19)^2 = 360