已知圆:x^2+y^2-4x-6y+12=0的圆心在点C,点A(3,5)

2025-12-25 05:23:53
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回答1:

1
圆C:(x-2)^2+(y-3)^2=1 设切点为P(x,y),
向量CP=(x-2,y-3),向量AP=(x-3,y-5)所以(x-2,y-3)(x-3,y-5)=0即(x-2)(x-3)+(y-3)(y-5)=0
y=3或x=7/5,y=14/5
即P(3,3)或P(7/5,14/5)
x=3或11 x-8y+7=02AOC的面积S=0.5