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急!F5 Equations of Circles 8q14
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(a) Since angle OAM = angle OBM = 90 degree. So OMAB is a cyclic quadrilateral ( angle in the same segment equal). (ii) So angle POB = angle AMP = 45 degree ( angle in the same segment). (bi) MB is the vertical line passing through B ( - 4, 0), so line MB is x = - 4 or x + 4 = 0 (bii) Since angle POB = 45 degree and angle PBO is a rt. angle, so PB = OB = 4 so P is ( - 4, 4) = center of circle. PB = radius of circle = 4 so equation of circle is (x + 4)^2 + (y - 4)^2 = 4^2 = 16
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急!F5 Equations of Circles 8q14
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請詳細步驟教我計下條 : 不要網址回答: 圖片參考：http://imgcld.yimg.com/8/n/HA05788109/o/20121213172934.jpg 答案 aii) 45度 bi) x+4=0 bii) (-4,4) (x+4)^2=(y-4)^2=16最佳解答:
(a) Since angle OAM = angle OBM = 90 degree. So OMAB is a cyclic quadrilateral ( angle in the same segment equal). (ii) So angle POB = angle AMP = 45 degree ( angle in the same segment). (bi) MB is the vertical line passing through B ( - 4, 0), so line MB is x = - 4 or x + 4 = 0 (bii) Since angle POB = 45 degree and angle PBO is a rt. angle, so PB = OB = 4 so P is ( - 4, 4) = center of circle. PB = radius of circle = 4 so equation of circle is (x + 4)^2 + (y - 4)^2 = 4^2 = 16
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