How do you write the general form given a circle that passes through the given points. P(-2, 0), Q(1, -3), R(-2, -6)?

1 Answer
Apr 15, 2016

General of equation of the circle would be x^2+y^2+4x+6y+4=0x2+y2+4x+6y+4=0

Explanation:

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=-1301(2)=33=1 and mid point of PQ is ((-2+1)/2,(0-3)/2)(2+12,032) 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=111=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=16(3)21=33=1 and mid point of QR is ((-2+1)/2,(-6-3)/2)(2+12,632) 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=-111=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=x12 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