Question

Find the equation of the circle circumscribing a triangle

Answer 1

General equation of a circle with centre (g,f) is x2+y2+2gx+2fy+c=0
Answer : x2+y2+122x−196y=0
Let the equations of the sides AB,BC and CA of the triangle be y=x+2----(1),3y=4x-----(2) and 2y=3x----(3) respectively.
On solving (1) & (2) we get,
3(x+2)=4x
⇒3x+6=4x
∴x=6 and y=8
Solving (2) & (3) we get,
x=0,y=0
Solving (3) & (1) we get,
2(x+2)=3x
⇒2x+4=3x
∴x=4 and y=6
Hence the vertices of the triangle are A(4,6),B(6,8) and C(0,0)
The circle x2+y2+2gx+2fy+c=0 passes through (4,6),(6,8) and (0,0)
∴8g+12f+c=−52⇒2g+3f=−13
12g+16f+c=−100⇒3g+4f+c=−50
c=0
2g+3f=−13----(4)
3g+4f=−50----(5)
(Eq(4)×3)6g+9f=−39
−(Eq(5)×2)6g+8f=−100
______________________________
⇒f=61
∴g=−98
c=0
Hence the equation of the required circle is x2+y2+122x−196y=0