Question

过P(5,4)作圆C:x^2+y^2-2x-2y-3=0的切线,切点分别为A,B,四边形PACB的面积是?

Answer 1

几何法求解.圆:(x-1)^2+(y-1)^2 = 5

可知 C(1,1); P(5,4) ; CP = 5; 圆C半径r = √5

于是 AP=PB=√(5^2-5) = 2√5

四边形PACB = r*(AP+BP)/2 = 10