Thursday

July 24, 2014

July 24, 2014

Posted by **John** on Sunday, May 5, 2013 at 5:53am.

- Analytic Geometry -
**Reiny**, Sunday, May 5, 2013 at 9:49amslope 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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