A circle has a center that falls on the line `y = 8/7x +2` and passes through `( 2 ,1 )` and `(3 ,9 )`. What is the equation of the circle?

`color(blue)("Method")`

1. Determine equation of line passing through BCD
2. Determine mid point of BD`->`C

3. Determine equation of the line passing through CA

4. Using simultaneous equation comparing `y=8/7x+2` to line
   `""`through AC determine point A (Centre of circle).

5. Determine equation of circle
   '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
   `color(blue)("Step 1 - Determine equation of line passing through BCD ")`

Gradient`" "->("change in y")/("change in x") =m=(9-1)/(3-2)=8`

So `y=mx+c" "->" "y=8x+c`

I chose line to pass through `P_1->B->(2,1)`

`=>1=8(2)+c" "-> c=1-16=-15`

So `color(blue)("BCD"->y=mx+c" "->" "y=8x+-15)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Step 2 - Determine mid point of BD"->"C ")`

`"C"-> "mean point "-> (x_1+x_2)/2" and " (y_1+y_2)/2`

`color(blue)("C"->(x,y)->(5/2 ,10/2) -> (5/2,5))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 3 - Determine equation of the line passing through CA")`

Now known: gradient `=-1/m = -1/8`
Now known: passes through the point C`->(5/2,5)`

Thus for this line `y=-1/mx+c" " ->" " 5=-1/8(5/2)+c`

`c=5 5/16 = 85/16`

`color(blue)(y=-1/mx+c" "->" "y=-1/8x+85/16)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Step 4 - Using simultaneous equation comparing"``color(blue)(y=8/7x+2" to line "y=-1/8x+85/16)`

`y=8/7x+2` ..........................(1)
`y=-1/8x+86/16` ..................(2)

Equation (1) - Equation (2) to eliminate y

`0=8/7x+1/8x+2-86/16`

`0=71/56x-84/16`

`color(blue)(x=84/16xx56/71 =4 41/142)` ................(3)

Substitute (3) into (1)

`y=8/7x+2" "->" "y=8/7(4 41/142)+2`

`color(blue)("y=6 64/71)`

`color(blue)("Centre of circle at "(x_c,y_c)" "->" "(4 41/142, 6 64/71)")`
`color(red)("Not very nice numbers!")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 5 - Determine equation of circle")`

If the circle is centred at the origin then the equation is

`r^2=x^2+y^2`

If the circle is offset then mathematically we transpose it back to the origin. So in this case, using step 4, we have:

`color(brown)( r^2=(x-4 41/142)^2+(y-6 64/71)^2 larr " equation of the circle")`

All we need to do now is determine the magnitude of r which is the distance AB

`=> r^2 =(x_c-x_1)^2+(y_c-y_1)`

`=> r^2 =(4 41/142-2)^2+(6 64/71-1)^2`

`color(red)("Switching do decimal as the numbers are getting silly!")`

`r~~6.329 ->6.33` to 2 decimal places.

`bar(color(blue)("| The equation of the circle is:"color(white)(......................)|)`
`underline(color(blue)(|=> 40.06~~(x-4 41/142)^2+(y-6 64/71)^2color(white)(.) |)`