Question 771638
Consider the general equation of a circle as {{{X^2 + Y^2 + DX + EY + F = 0}}}.

given: points ({{{1}}},{{{1}}}) ({{{2}}},{{{-1}}}), ({{{2}}},{{{3}}})

Using the given three points we derive our equation :

From the point ({{{1}}},{{{1}}}) we get the equation 

{{{1+ 1 + D + E + F = 0}}} or

{{{D + E + F+ 2 = 0}}} ........  equation 1

From the point ({{{2}}},{{{-1}}}) we get 

{{{4 + 1 + 2D -E + F = 0}}} or

{{{2D -E + F+5 = 0}}}........  equation 2


From the point ({{{2}}},{{{3}}}) we get 

{{{4 + 9 +2D + 3E + F = 0}}} or 

{{{2D + 3E + F+13 = 0}}}.......  equation 3


We now have a system of three equations :


{{{D + E + F+ 2 = 0}}} ........  equation 1

{{{2D -E + F+5 = 0}}}........  equation 2

{{{2D + 3E + F+13 = 0}}}.......  equation 3


We now use elimination method to find the value of {{{D}}},{{{ E}}} and {{{F}}}.

Using equation 1 and eq. 2 to eliminate {{{F}}}, we do this by subtracting equation 2 from eq. 1 :

{{{D + E + F+ 2-( 2D -E + F+5 )= 0}}}

{{{D + E + F+ 2-2D +E - F-5 = 0}}}

{{{2E -2D-3 = 0}}}..... let this be equation 4


Using Equation 2 and eqn 3, if we subtract eq. 3 from eq. 2 we can eliminate {{{F}}} and {{{D}}}:

{{{2D -E + F+5 -(2D + 3E + F+13 )= 0}}}

{{{2D -E + F+5 -2D-3E - F-13 = 0}}}

{{{-E +5 -3E -13 = 0}}}

{{{-4E  -8 = 0}}}

{{{-E  -2 = 0}}} => {{{highlight(E =-2)}}}

go back to eq.4 and plug in {{{E}}}

{{{2(-2) -2D-3 = 0}}}.......solve for {{{D}}}

{{{-4 -3 = 2D}}}

{{{-7 = 2D}}}

{{{-7 /2= D}}}

{{{highlight(D=-3.5)}}}

go back to eq.1 and plug in {{{E}}} and {{{D}}} and solve for {{{F}}}

{{{-3.5-2 + F+ 2 = 0}}} ........  equation 1

{{{-3.5 + F = 0}}}

 {{{highlight(F = 3.5)}}}

We get this general equation by substituting 

{{{D = -3.5}}}, {{{E = -2}}} and {{{F = 3.5}}} to the general equation of the circle 

{{{X^2 + Y^2 + DX + EY + F = 0}}}:

{{{X^2 + Y^2-3.5X -2Y + 3.5 = 0}}}