How do you write the equation for a circle through (-2,4), (6,0) and (1,5)?

1 Answer
Aug 25, 2016

We assume the following result :

Result : If a circle #S' : x^2+y^2+2gx+2fy+c=0, "and, a line" L : #

#lx+my+n=0 " intersect, then" S'+lambdaL=0, lambda in RR,#

represents a circle that passes through their points of intersection.

Consider a circle #S'# having diametrically opposite pts.

#(-2,4) and (1,5)#. Then,

#S' : (x+2)(x-1)+(y-4)(y-5)=0,#, or,

# S' : x^2+y^2+x-9y+18=0#

The Eqn. of the line #L# through these pts.
#L: det|(x,y,1),(-2,4,1),(1,5,1)|=0, i.e., L : -x+3y-14=0#.

We observe that the Reqd. Circle #S# passes through the pts. of intersection of circle #S'# and line #L#. Hence, by the above Result,

# S : S'+lambdaL=0 :#, i.e.,

# S : x^2+y^2+x-9y+18+lambda(-x+3y-14)=0, lambda in RR#

The pt. #(6,0) in S#,

#rArr 36+0+6-0+18+lambda(-6+0-14)=0#.

#rArr 60-20lambda=0 rArr lambda=3#. Hence,

# S : x^2+y^2-2x-24=0#.

Enjoy Maths.!