已知圆满足:①截y轴所得弦长为2;②被x轴分成两段圆弧,其弧长

 已知圆满足:y轴所得弦长为2x轴分成两段圆弧,其弧长的比为3∶1圆心到直线lx2y0的距离为,求该圆的方程.

答案

解:设圆P的圆心为P(ab),半径为r,则点Px轴、y轴的距离分别为|b||a|.

由题设知圆Px轴所得劣弧所对圆心角为90°,知圆Px轴所得的弦长为r.

2|b|r,得r22b2

又圆Py轴所截得的弦长为2,由勾股定理得r2a21,得2b2a21.

又因为P(ab)到直线x2y0的距离为

所求圆的方程是(x1)2(y1)22,或(x1)2(y1)22.

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