Question #1347
Find the equation of the circle inscribed the triangle determind by the lines:2x-3y+21=0;3x-2y-6=0 and 2x+3y+9=0.
Expert's answer
Firstly, we have to find the points of intersection of thelines:
1. 2x-3y+21=0 and 3x-2y-6=0 :
(12, 15)
2. 2x-3y+21=0 and 2x+3y+9=0 :
(-7.5, 2)
3. 3x-2y-6=0 and 2x+3y+9=0 :
(0; -3).
As the equation of the circle is
R2 = (x-a)2+ (y-b)2,
we have the final system of equations for a and b:
(12-a)2 + (15-b)2 = (-7.5-a)2+ (2 – b)2
(12-a)2 + (15-b)2 =a2 + (-3-b)2
144 + 225 – 24a -30b = 56.25 + 4 + 15a – 4b;
144+ 225 – 24a -30b = 9 – 6b.
308.75 = 39a + 26b;
360 = 24a +24b.
(-6.25, 21.25)
Thus R2 = (12+ 6.25)2 + (15-21.25)2= 372.125
Answer: (x+6.25)2 + (y-21.25)2=372.125
1. 2x-3y+21=0 and 3x-2y-6=0 :
(12, 15)
2. 2x-3y+21=0 and 2x+3y+9=0 :
(-7.5, 2)
3. 3x-2y-6=0 and 2x+3y+9=0 :
(0; -3).
As the equation of the circle is
R2 = (x-a)2+ (y-b)2,
we have the final system of equations for a and b:
(12-a)2 + (15-b)2 = (-7.5-a)2+ (2 – b)2
(12-a)2 + (15-b)2 =a2 + (-3-b)2
144 + 225 – 24a -30b = 56.25 + 4 + 15a – 4b;
144+ 225 – 24a -30b = 9 – 6b.
308.75 = 39a + 26b;
360 = 24a +24b.
(-6.25, 21.25)
Thus R2 = (12+ 6.25)2 + (15-21.25)2= 372.125
Answer: (x+6.25)2 + (y-21.25)2=372.125
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#339914 on Dec 2023
#339914 on Dec 2023