$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 5:\\ Ta\ có\ \Delta :\ y-3=-\frac{1}{2}( x-1) \Leftrightarrow y=-\frac{1}{2} x+\frac{7}{2} \ nhận\ \overrightarrow{n_{1}} \ ( 1;2) \ là\ vtpt\\ \Delta ':\ 2x-3y-1=0\ hay\ y=\frac{2x-1}{3} \ nhận\ \overrightarrow{n_{2}}( 2;-3) \ là\ vtpt\\ 1.\ \Delta \ và\ \Delta '\ giao\ nhau\ \Leftrightarrow -\frac{1}{2} x+\frac{7}{2} =\frac{2x-1}{3} \ \Leftrightarrow x=\frac{23}{7} \ thay\ vào\ \Delta \ ta\ có:y=\frac{13}{7}\\ Ta\ có\ cos\ \widehat{\Delta ,\Delta '} =|\frac{\overrightarrow{n_{1}} .\overrightarrow{n_{2}}}{\overrightarrow{|n_{1}} ||\overrightarrow{n_{2} |}} |=\frac{|1.2+2.( -3) |}{\sqrt{1^{2} +2^{2}}\sqrt{2^{2} +( -3)^{2}}} =\frac{2\sqrt{2}}{5}\\ 2.\ do\ d_{1} //\Delta \Rightarrow \overrightarrow{{n_{d}}_{1}} //\overrightarrow{n_{1}}\\ d_{1} \ đi\ qua\ M( 2;-5) \ nhận\ \overrightarrow{n_{1}} là\ vtpt\ có\ PT:\ 1( x-2) +2( y+5) =0\ hay\ x+2y+8=0\\ 3.\ d_{2} \perp \Delta '\Rightarrow \overrightarrow{{n_{d}}_{2}} \perp \overrightarrow{n_{2}}\\ d_{2} \ đi\ qua\ N( 4;-8) \ nhận\ \overrightarrow{n_{2}}( 2;-3) \ là\ vtcp\ hay\ \overrightarrow{v_{2}}( 3;2) \ là\ vtpt,\ có\ PTĐT:\\ 3( x-4) +2( y+8) =0\ hay\ 3x+2y+4=0\\ 4.\ K\in \Delta \Rightarrow \ K\left( a;-\frac{1}{2} a+\frac{7}{2}\right)\\ d( K,\Delta ') =\frac{|2a-3.\left( -\frac{1}{2} a+\frac{7}{2}\right) -1|}{\sqrt{2^{2} +( -3)^{2}}} =4\\ \Leftrightarrow |\frac{7}{2} a-\frac{23}{2} |=4\sqrt{13} \Leftrightarrow \left(\frac{7}{2} a-\frac{23}{2}\right)^{2} =208\Leftrightarrow \frac{49}{4} a^{2} -\frac{161}{2} a+\frac{529}{4} -208=0\\ \Leftrightarrow a=\frac{23+8\sqrt{13}}{7} \ hoặc\ a=\frac{23-8\sqrt{13}}{7}\\ Vậy\ K\left(\frac{23+8\sqrt{13}}{7} ;\frac{13-4\sqrt{13}}{7}\right) \ hoặc\ K\left(\frac{23-8\sqrt{13}}{7} ;\ \frac{13+4\sqrt{13}}{7}\right)\\ 5.\ Gọi\ I\left( b;\frac{2b-1}{3}\right) \in \Delta '\ là\ tâm\ đường\ tròn\ ( C)\\ Do\ ( C) \ tiếp\ xúc\ với\ \Delta \ và\ R=5\ \Rightarrow \ d( I;\Delta ) =5\\ \Leftrightarrow \frac{|\frac{1}{2} b+\frac{2b-1}{3} -\frac{7}{2} |}{\sqrt{1+( 2)^{2}}} =5\Leftrightarrow |\frac{7}{6} b-\frac{23}{6} |=5\sqrt{5} \Leftrightarrow \left(\frac{7}{6} b-\frac{23}{6}\right)^{2} =\left( 5\sqrt{5}\right)^{2}\\ \Leftrightarrow b=\frac{23+30\sqrt{5}}{7} \ hoặc\ b=\frac{23-30\sqrt{5}}{7}\\ \Rightarrow I\left(\frac{23+30\sqrt{5}}{7} ;\frac{13+20\sqrt{5}}{7}\right) \ hoặc\ I\left(\frac{23-30\sqrt{5}}{7} ;\frac{13-20\sqrt{5}}{7}\right) \ \\ Vậy\ ( C) \ có\ PT\ đường\ tròn\ là\ ( C) :\ \left( x-\frac{23+30\sqrt{5}}{7}\right)^{2} +\left( y-\frac{13+20\sqrt{5}}{7}\right)^{2} =25\\ hoặc\ \left( x-\frac{23-30\sqrt{5}}{7}\right)^{2} +\left( y-\frac{13-20\sqrt{5}}{7}\right)^{2} =25\\ Bài\ 6:\\ Ta\ có\ AB=\sqrt{( 1-0)^{2} +( 5+2)^{2}} =5\sqrt{2\ }\\ \ \ \ \ \ \ \ \ \ \ AC=\sqrt{( 1-3)^{2} +( 5-7)^{2}} =2\sqrt{2\ }\\ \ \ \ \ \ \ \ \ \ \ BC=\sqrt{( 3-0)^{2} +( 7+2)^{2}} =3\sqrt{10\ }\\ Có\ p=\frac{AB+BC+AC}{2} =\frac{5\sqrt{2\ } +2\sqrt{2\ } +3\sqrt{10\ }}{2} =\frac{7\sqrt{2\ } +3\sqrt{10\ }}{2}\\ Ta\ có\ S_{ABC} =\frac{AB.AC.BC}{4R} \Rightarrow R=\frac{AB.AC.BC}{4S_{ABC}} =\frac{AB.AC.BC}{4\sqrt{p( p-AB)( p-AC)( p-BC)}} \end{array}$