Oddly defined triangle

Let (x,y) & (z,t) be the other two vertices. Then we can set up the following equations x=-z, y+t=-4,29=(x+2)²+(y+1)²,29=(z+2)²+(t+1)² and ,2A=x(4-t)+z(y-4) where A is the area of the triangle. A=12√( $\frac{6}{5}$ ) and thus a+b+c=12+6+5=23

$\color{#BA33D6}{We~shift~ the~axes~to~(-2, -1).\\ So~ the~ new~coordinates ~are~A(2,5);~~O(0,0);~~G(2,1).\\ ~~~\\ ~~~\\ Let~H~be~the~orthocenter. Then~we~know~OH=3*OG.~~~~~ So~\color{#3D99F6}{H(6,3)}.\\ ~~~\\ ~~~\\ Let~M~be ~the~the ~foot~of~altitude~from~A.~It~passes~through~H.\\ So~Y_{AM}=\dfrac{5-3}{3-6}*(X-2)+5=-\frac X 2+6..........(1)\\ \implies~slope~of~BC~is~2.........(2)\\ ~~~\\ Circum\bigcirc~is~x^2+y^2=2^2+5^2=29. ~O(0,0)~as~center.......(3)\\ ~~~\\ ~~~\\ Let~AM~extended~meet~\bigcirc~at~N. \\ By ~(1)~and~(3)~~X_N^2+(-\frac 1 2 X_N+6)^2=29.\\ Solving~quadratic~in~X_N,~~we~get,~\dfrac {14} 5~~and~~2.\\ We~ know~2=X_A, ~so~X_N=\dfrac {14} 5.\\ From~(1)~~get~Y_N=\dfrac {23} 5.\implies~\color{#3D99F6}{N \Big(\dfrac {14} 5,\dfrac {23} 5 \Big)} .\\ ~~~\\ ~~~\\ We~also~know~that ~N~is ~the~ reflection~of~ H~wrt~BC.\\ \implies~M~is~the~midpoint~of~HN.~~So~\color{#3D99F6}{M\Big(\dfrac {22} 5,~\dfrac {19} 5\Big) }.\\ ~~~\\ ~~~\\ By~(2),~~Y_{BC}=2\Big(X- \dfrac {22} 5\Big)+\dfrac {19} 5=2X+5 .\\ So~chord~BC~is~at~\bot~distance~of~\dfrac{ 0+0+ 5}{\sqrt{2^2+1^2}}=\sqrt5~from ~O.\\ So~\frac 1 2 BC=\sqrt{29-(\sqrt5)^2}=\sqrt{24}=\color{#3D99F6}{2\sqrt6} ..........(4)\\ ~~~\\ ~~~\\ Altitude~AM=\sqrt{(X_A-X_M)^2+(Y_A-Y_M)^2}=\color{#3D99F6}{6*\dfrac 1 {\sqrt5} }.\\ ~~~\\ ~~~\\ By~(4)~~Area~\Delta~ABC=\frac 1 2 BC*AM=2*\sqrt6*\Big (6 *\dfrac 1 {\sqrt5} \Big ).\\ ~~~\\ ~~~\\ Area~=~12*\sqrt{\dfrac 6 5}=a*\sqrt{\frac b c}.\\ a+b+c=12+6+5=\color{#D61F06}{23}. } \\ ~~~\\ ~~~\\$

$\color{#20A900}{Summary:- ~~~~We~ shifted ~the~origin~to~Circum\bigodot,~O(-2,-1).\\ Found~orthocenter~H~on~Euler~ Line ~from~given~O~and~G.\\ Found~reflection~of~H~wrt~BC~as~N~where~Extended~AH~meet~Circum\bigcirc.\\ Obtain~equation~of~c~chord~and~side~0f~\Delta~ABC,~~BC~passing~through~M~midpoint~of~HN.\\ Found~\bot~distance~of~BC~from~Circum\bigodot~and~hence~half~chord~length.\\ Found ~altitude~length~AM. Thus~area~of~\Delta~ABC.}$

$OR$

$\color{#EC7300}{Second~approach:-\\ We~ shifted ~the~origin~to~Circum\bigodot,~O(-2,-1).\\ Find~orthocenter~H~on~Euler~ Line ~from~given~O~and~G.\\ Find~D~of~median~AD, by~extending~AG~to~D,~~AO=2OD.\\ Get~slope~of~BC~as~\bot~to~AH.\\ Find~\bot~distance~of~chord~BC~from~O.~Hence~\frac 1 2 BC.\\ Extend~AH~to~meet~BC~at~M. ~Find~length~AD.\\ Area= \frac 1 2 BC*AD.}\\$