What is the standard form of the equation of a circle passing through (0,8), (5,3) and (4,6)?

1 Answer
Nov 29, 2016

Answer:

I have taken you to a point where you should be able to take over.

Explanation:

#color(red)("There may be an easier way of doing this")#

The trick is to manipulate these 3 equations in such a way that you end up with 1 equation with 1 unknown.

Consider the standard form of #(x-a)^2+(y-b)^2=r^2#

Let point 1 be #P_1->(x_1,y_1) =(0,8)#
Let point 2 be #P_2->(x_2,y_2)=(5,3)#
Let point 3 be #P_3->(x_3,y_3)=(4,6)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For #P_1 ->(x_1-a)^2+(y_1-b)^2=r^2#

#(0-a)^2+(8-b)^2=r^2#

#a^2+64-16b+b^2=r^2#...............Equation(1)
............................................................................................................
For #P_2->(x_2-a)^2+(y_2-b)^2=r^2#

#(5-a)^2+(3-b)^2=r^2#

#25-10a+a^2+9-6b+b^2=r^2#

#a^2-10a+34-6b+b^2=r^2#............Equation(2)
..........................................................................................................

For #P_3->(x_3-a)^2+(y-b)^2=r^2#

#(4-a)^2+(6-b)^2=r^2#

#16-8a+a^2+36-12b+b^2=r^2#

#a^2-8a+52-12b+b^2=r^2#...........Equation(3)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Lets see where this gets us!

Equation(3) - Equation(2)

#a^2-8a-12b+b^2+52=r^2#
#ul(a^2-10a-6b+b^2+34=r^2) larr" subtract"#
#0" "+2a -6b+0+18=0#

#2a-6b+18=0# ...........................Equation(4)

#=>a=(6b-18)/2 = 3b-9#

#color(brown)("we can now substitute for "a)##color(brown)("in equations 1 and 2 and solve for "b)#

#equation(1)=r^2=equation(2)#

#a^2-16b+b^2" " =" " a^2-10a-6b+b^2+34#

#cancel(a^2)-16b+cancel(b^2)" " =" " cancel(a^2)-10a-6b+cancel(b^2)+34#

Substituting for #a#

#-16b" "=" "-10(3b-9)-6b+34#

#-16b" "=" "-30b+90-6b+34#

#-16b" "=" "-36b+124#

#" "color(green)(ul(bar(|" "b=124/20=31/5" "|))#

#color(red)("I will let you take it on from this point")#