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由已知可得,內切圓可轉換為(x-2)^2+(y-k)^2=k^2,即,該圓必過點(2,k),設圓心座標為O(2,k),則OB直線的斜率為K1=k/4,設頂點為A,則根據已知條件中的內切圓,所以角ABC被線段OB平分,則OA的斜率可由二倍角正切計算…
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