Question

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

Answer 1

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.