How do you use the information provided to write the equation of each circle:Three points on the circle are (-15, -6), (-19,-4), and (-17, -6)?

1 Answer
Feb 6, 2017

Answer:

Equation of only circle is #x^2+y^2+32x+6y+255=0#

Explanation:

As there are three points on the circle are #(-15, -6)#, #(-19,-4)# and #(-17, -6)#, there is only one circle and the center of circle will lie on perpendicular bisector of any two sides.

Let us find the equation of perpendicular bisector of points joining #(-15, -6)# and #(-19,-4)#.

As a point on a perpendicular bisector is equidistant from #(-15, -6)# and #(-19,-4)#, its equation will be

#(x+15)^2+(y+6)^2=(x+19)^2+(y+4)^2#

or #x^2+30x+225+y^2+12y+36=x^2+38x+361+y^2+8y+16#

i.e. #30x-38x+12y-8y=377-261#

or #-8x+4y=116# or #2x-y=-29# ....................(1)

Similarly equation of perpendicular bisector of points joining #(-17, -6)# and #(-19,-4)# is

#(x+17)^2+(y+6)^2=(x+19)^2+(y+4)^2#

or #x^2+34x+289+y^2+12y+36=x^2+38x+361+y^2+8y+16#

i.e. #34x-38x+12y-8y=377-325#

or #-4x+4y=52# or #x-y=-13# ....................(2)

Subtracting equation (2) from (1), we get

#x=-16# and hence putting this in (2), we get #y=-3#

Hence center of the circle is #(-16,-3)# and square of the radius is square of the distance of center from say #(-17,-6)# i.e.

#(-16+17)^2+(-3+6)^2=1+9=10#

Hence equation of circle is

#(x+16)^2+(y+3)^2=10#

or #x^2+32x+256+y^2+6y+9=10#

or #x^2+y^2+32x+6y+255=0#
graph{x^2+y^2+32x+6y+255=0 [-20.645, -10.645, -6.88, -1.88]}