[tex](x+5)^2+(y-12)^2=14^2<=>x^2+y^2+10x-24y+169=196=>x^2+y^2=-10x+24y+27=-10(x+5)+24(y-12)+365[/tex]
xét [tex]|-10(x+5)+24(y-12)|\leq \sqrt{((-10)^2+24^2)((x+5)^2+(y-12)^2)}=364=>-364\leq -10(x+5)+24(y-12)\leq 364[/tex]
=>[tex]x^2+y^2\geq -364+365=1[/tex]
dấu bằng xảy ra khi -10/24=(x+5)/(y-12)