How do you find the equation for the circle where A(7,-6),B(9,2) and C(1,4) are points on a circle?

1 Answer
George C.
Aug 6, 2016

#(x-4)^2+(y+1)^2 = 34#

Explanation:

Notice that #vec(AB) = (2,8)# and #vec(BC) = (-8,2)# are perpendicular to one another and of the same length.

So #ABC# is an isosceles right angled triangle. Together with the point #D(-1, -4)# the points #A,B,C,D# form the vertices of a square with centre at the midpoint of #AC#:

#E = ((7+1)/2, (-6+4)/2) = (4, -1)#

The radius of the circumscribing circle is the distance between the centre #E# and any of #A,B,C,D# say #C#:

#r = sqrt((1-4)^2+(4-(-1))^2) = sqrt(3^2+5^2) = sqrt(34)#

The equation of a circle with centre #(h,k)# and radius #r# may be written:

#(x-h)^2+(y-k)^2 = r^2#

Hence the equation of our circle may be written:

#(x-4)^2+(y-(-1))^2 = 34#

or slightly easier to read as:

#(x-4)^2+(y+1)^2 = 34#

graph{((x-4)^2+(y+1)^2-34)((x-4)^2+(y+1)^2-0.03)((x-7)^2+(y+6)^2-0.08)((x-9)^2+(y-2)^2-0.08)((x+1)^2+(y+4)^2-0.08)((x-1)^2+(y-4)^2-0.08) = 0 [-14.67, 25.33, -10.44, 9.56]}

Related questions
See all questions in Identify Critical Points
Impact of this question
1192 views around the world
You can reuse this answer
Creative Commons License