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        圓x2+y2-4x-4y-10=0上的店到直線x+y-14=0的最大距離和

        圓A:x^2+y^2-4x-4y-10=0
        (x-2)^2+(y-2)^2=2
        圓心A(2,2) 半徑√2
        l1:y=x 垂直于l2:x+y-14=0
        l1,l2交于M(7,7)
        AM=(7-2)√2
        AM直線交圓A于PQ
        MP=5√2-√2=4√2
        MQ=5√2+√2=6√2

        Hash:4ffa844ab262aea80b70a381f6a5bc10b6e41769

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