We are given three points P(-2,0)P(−2,0), Q(1,-3)Q(1,−3) and R(-2,-6)R(−2,−6).
The perpendicular bisectors of any two lines joining these points will intersect at the center of the circle passing through these points.
Slope of line joining PQ is (-3-0)/(1-(-2))=-3/3=-1−3−01−(−2)=−33=−1 and mid point of PQ is ((-2+1)/2,(0-3)/2)(−2+12,0−32) or (-1/2,-3/2)(−12,−32). Hence its perpendicular bisector will pass through (-1/2,-3/2)(−12,−32) and have a slope of -1/-1=1−1−1=1.
Hence equation of perpendicular bisector of PQ is y-(-3/2)=1xx(x-(-1/2))y−(−32)=1×(x−(−12)) or y+3/2=x+1/2y+32=x+12 or x=y+1x=y+1.
Slope of line joining QR is (-6-(-3))/(-2-1)=-3/-3=1−6−(−3)−2−1=−3−3=1 and mid point of QR is ((-2+1)/2,(-6-3)/2)(−2+12,−6−32) or (-1/2,-9/2)(−12,−92). Hence its perpendicular bisector will pass through (-1/2,-9/2)(−12,−92) and have a slope of -1/1=-1−11=−1.
Hence equation of perpendicular bisector of PQ is y-(-9/2)=-1xx(x-(-1/2))y−(−92)=−1×(x−(−12)) or y+9/2=-x-1/2y+92=−x−12 or x+y+5=0x+y+5=0.
Solving x=y+1x=y+1 and x+y+5=0x+y+5=0 gives us (-2,-3)(−2,−3), which is center OO and radius will be its distance with say P(-2,0)P(−2,0), which is 33
Hence equation of circle will be (x-(-2))^2+(y-(-3))^2=3^2(x−(−2))2+(y−(−3))2=32
or (x+2)^2+(y+3)^2=9(x+2)2+(y+3)2=9 or x^2+4x+4+y^2+6y+9=9x2+4x+4+y2+6y+9=9 and simplifying it general for of equation of circle would be x^2+y^2+4x+6y+4=0x2+y2+4x+6y+4=0