<prompt> Mduara C wenye kipenyo cha 2 una kipenyo overlineAB. Mduara D una mduara wa ndani ili kuzunguka C kwa A. Mduara E ni sanjari ya ndani ili kuzunguka C, mduara wa nje kuzunguka D, na tanjent hadi overlineAB. Radi ya mduara D ni mara tatu ya kipenyo cha duara E, na inaweza kuandikwa katika muundo sqrtm-n, ambapo m na n ni nambari kamili chanya. Tafuta mn.  <response>: asy kuagiza grafu; ukubwa 7.99cm; labelscalefactor halisi 0.5; kalamu dps upana wa mstari 0.7 fontsize 10; defaultpendps; kalamu dotstyle nyeusi; xmin halisi 4.087153740193288, xmax 11.08175859031552, ymin -4.938019122704778, ymax 1.194137062512079; choramduara7.780000000000009,-1.320000000000002, 2.000000000000000; choramduara7.271934046987930,-1.319179731427737, 1.491933384829670; choramduara9.198812158392690,-0.8249788848962227, 0.4973111282761416; kuchora 5.780002606580324,-1.316771019595571--9.779997393419690,-1.323228980404432; kuchora9.198812158392690,-0.8249788848962227--9.198009254448635,-1.322289365031666; kuchora 7.271934046987930,-1.319179731427737--9.198812158392690,-0.8249788848962227; kuchora9.198812158392690,-0.8249788848962227--7.7800000000000009,-1.32000000000002; nukta7.780000000000009,-1.320000000000002, mtindo wa nukta; leboC, 7.707377218471464,-1.576266740352400, NE labelscalefactor; nukta 7.271934046987930,-1.319179731427737, mtindo wa nukta; leboD, 7.303064016111554,-1.276266740352400, NE labelscalefactor; nukta 9.198812158392690,-0.8249788848962227, mtindo wa nukta; leboE, 9.225301294671791,-0.7792624249832147, NE labelscalefactor; nukta 9.198009254448635,-1.322289365031666, mtindo wa nukta; leboF, 9.225301294671791,-1.276266740352400, NE labelscalefactor; nukta 9.779997393419690,-1.323228980404432, mtindo wa nukta; leboB, 9.810012253929656,-1.276266740352400, NE labelscalefactor; nukta 5.780002606580324,-1.316771019595571, mtindo wa nukta; leboA, 5.812051070003994,-1.276266740352400, NE labelscalefactor; klipuxmin,ymin--xmin,ymax--xmax,ymax--xmax,ymin--mzunguko; asyKwa kutumia mchoro ulio hapo juu, acha kipenyo cha D kiwe 3r, na kipenyo cha E kiwe r. Kisha, EFr, na CE2-r, kwa hivyo nadharia ya Pythagorean katika pembetatu CEF inatoa CFsqrt4-4r. Pia, CDCA-AD2-3r, hivyoDFDCCF2-3rsqrt4-4r.Tukikumbuka kuwa DE4r, sasa tunaweza kutumia Pythagorean theorem katika pembetatu DEF kupata2-3rsqrt4-4r2r216r2.Kutatua quadratic hii kwa kiasi fulani kunachosha, lakini masharti ya mara kwa mara yanaghairi, kwa hivyo hesabu sio mbaya. Kutatua kunatoa 3rsqrt240-14 kwa jibu la mwisho la 254.Tambua kwamba C, E na hatua ya kukunja mduara C kwa mduara E itakuwa sawa kwa sababu C na E hukatiza mstari wa tanjiti kwenye pembe ya kulia, ikimaanisha kuwa lazima ziwe kwenye mstari mmoja. 