“微信扫一扫”进入题库练习及模拟考试
已知圆
设点
参考答案:\({(x+2{)}^{2}}+{(y+1{)}^{2}}\)可看作点\(P\left( {x,y} \right)\)与点\(\left( { - 2, - 1} \right)\)距离的平方,
圆心\(C\left( {2,2} \right)\)与点\(\left( { - 2, - 1} \right)\)的距离\(d = \sqrt {{{\left( {2 + 2} \right)}^2} + {{\left( {2 + 1} \right)}^2}} = 5 > 3\),
则点\(\left( { - 2, - 1} \right)\)在圆\( C\)外,
故点\(\left( { - 2, - 1} \right)\)到圆\( C\)上点的距离最大值为\(d + R = 8\),距离最小值为\(d - R = 2\).
所以,\({(x + 2)^2} + {(y + 1)^2}\)的最大值为\( 64\),最小值为\( 4\).
解析: