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)(.) |)#