What is the equation for the circle which passes through (3;4) (-4;3) (5;0)?

1 Answer
Jul 19, 2018

I will start you off

Explanation:

#color(blue)("Building the initial condition model")#

Let some constant related to #x# be #K_x#
Let some constant related to #y# be #K_y#

Let the radius be #r#

#color(green)("Then as a general equation we have:")#

#color(green)((x-K_x)^2+(y-K_y)^2=r^2" "......................Equation(1))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For point 1 #->P_1 ->(-4,3)" "Equation(1)# becomes:

#(-4-K_x)^2+(3-K_y)^2=r^2#

#(K_x)^2+8K_x+16+(K_y)^2-6K_y+9=r^2#

#color(brown)((K_x)^2+(K_y)^2+8K_x-6K_y+25=r^2color(white)("ddd")Eqn(1_a))#
......................................................................................

For point 2 #->P_2->(3,4)" "Equation(1)# becomes:

#(3-K_x)^2+(4-K_y)^2=r^2#

#(K_x)^2-6K_x+9+(K_y)^2-8K_y+16=r^2#

#color(brown)((K_x)^2 +(K_y)^2-6K_x-8K_y+25=r^2color(white)("ddd")Eqn(1_b) )#

...........................................................................
For point 3 #->P_2->(3,4)" "Equation(1)# becomes:

#(5-K_x)^2+(0-K_y)^2=r^2#

#(K_x)^2 -10K_x+25+(K_y)^2=r^2 #

#color(brown)((K_x)^2+(K_y)^2-10K_x+25=r^2color(white)("ddd") Eqn(1_c)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#

#color(brown)((K_x)^2+(K_y)^2+8K_x-6K_y+25=r^2color(white)("ddd")Eqn(1_a))#
#color(brown)((K_x)^2 +(K_y)^2-6K_x-8K_y+25=r^2color(white)("ddd")Eqn(1_b) )#
#color(brown)((K_x)^2+(K_y)^2-10K_xcolor(white)("ddddd")+25=r^2color(white)("ddd") Eqn(1_c)#

Now you manipulate these to gradually isolate the unknowns.
3 unknowns and 3 equations so solvable.