Question

find the equation of the circle passing through the three non collinear points (1,1), (2,-1) and (3,2).​

Answer 1

Answer:  x²+y²-5x-y-4=0

Step-by-step explanation:

It passes through(1 ,1)          let the equ  of circle    x²+y²+2gx+2fy+c=0    (A)

2g+2f+c=-2 equ  (1)

(2,-1) then 4g-2f+c= -5   equ (2)

(3,2) then  6g+4f+c= -13   equ (3)

solving(1) and (2) we get   -2g +4f=3  equ (4)

again (2)and (3)    -2g -6f=8  equ (5)

solving (4) &(5)     f=-1/2

sub f= -1/2 in  equ (4)   we get   g= -5/2

sub f &g in  equ (1)    c= -4

sub f,g&c in (A) we get  x²+y²-5x -y-4=0

Answer 2

Answer:

Step-by-step explanation:

this can be solved by 4 methods

i will solve first by perpendicular bisector method

perpendicular bisector of 1,1 and 2,-1 = 2x+y-3=0

perpendicular bisector of 1,1 and 3,2 = 2x-4y+2=0

point of intersection will be centre and centre to one of the coordinate distance will be radius