標題:
amaths circle
Find the coordinates of the centre of the circle(s) that passes through the point(1,-14) and touches the following 2 circles externally. x^2 + y^2 = 1 and (x-10)^2 + y^2 = 16. Answer to 3 sig. fig.
最佳解答:
The circle: x2 + y2 = 1 centre at O(0, 0) and radius = 1 (x-10)2 + y2 = 16. centre at Q(10, 0) and radius = √16 = 4 Let the centre of the required centre be C(a,b) with radius r, and the circle passes through the point P(1, -14). 圖片參考：http://i277.photobucket.com/albums/kk75/uncle_michael/080708_mat01.jpg?t=1215504581 CO = r + 1 √[(a - 0)2 + (b - 0)2] = r + 1 r = -1 + √(a2 + b2) ...... (1) CQ = r + 4 √[(a - 10)2 + (b - 0)2] = r + 4 r = -4 + √(a2 + b2 - 20a + 100) ...... (2) CP = r √[(a - 1)2 + (b + 14)2] = r r = √(a2 + b2 - 2a + 28b + 197) ...... (3) (1) = (2) -1 + √(a2 + b2) = -4 + √(a2 + b2 - 20a + 100) 3 + √(a2 + b2) = √(a2 + b2 - 20a + 100) 9 + a2 + b2 + 6√(a2 + b2) = a2 + b2 - 20a + 100 6√(a2 + b2) = -20a + 91 ...... (4) (1) = (3) -1 + √(a2 + b2) = √(a2 + b2 - 2a + 28b + 197) [-1 + √(a2 + b2)]2 = [√(a2 + b2 - 2a + 28b + 197)]2 1 + a2 + b2 - 2√(a2 + b2) = a2 + b2 - 2a + 28b + 197 -2√(a2 + b2) = -2a + 28b + 196 6√(a2 + b2) = 6a - 84b - 588 ...... (5) (4) = (5) - 20a + 91 = 6a - 84b - 588 84b = 26a - 679 b = (26a - 679)/84 ...... (6) Subst. (6) into (4) 6√{a2 + [(26a - 679)/84]2} = -20a + 91 36{a2 + (26a - 679)2/7056} = (-20a + 91)2 36[7056a2 + (26a - 679)2] = 7056 (-20a + 91)2 36[7056a2 + (26a - 679)2] = 7056 (-20a + 91)2 [7056a2 + (26a - 679)2] = 196(-20a + 91)2 7056a2 + 676a2 - 35308a + 461041 = 196(400a2 - 3640a + 8281) 7056a2 + 676a2 - 35308a + 461041 = 78400a2 - 713440a + 1623076 70668a2 - 678132a + 1162035 = 0 a = [678132 √(6781322 - 4 x 70668 x 1162035)] / (2 x 70668) Hence, a = 7.362 ororor a = 2.233 When a = 7.362, subst. into (6) b = (26 x 7.362 - 679)/84 b = -5.805 (Rejected because the circle touches internally the circle with centre Q.) When a = 2.233, subst. into (6) b = (26 x 2.233 - 679)/84 b = -7.392 Ans: The centre of the required circle is (2.233, -7.392) =
其他解答:960E6207E048F01D
amaths circle
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發問:Find the coordinates of the centre of the circle(s) that passes through the point(1,-14) and touches the following 2 circles externally. x^2 + y^2 = 1 and (x-10)^2 + y^2 = 16. Answer to 3 sig. fig.
最佳解答:
The circle: x2 + y2 = 1 centre at O(0, 0) and radius = 1 (x-10)2 + y2 = 16. centre at Q(10, 0) and radius = √16 = 4 Let the centre of the required centre be C(a,b) with radius r, and the circle passes through the point P(1, -14). 圖片參考：http://i277.photobucket.com/albums/kk75/uncle_michael/080708_mat01.jpg?t=1215504581 CO = r + 1 √[(a - 0)2 + (b - 0)2] = r + 1 r = -1 + √(a2 + b2) ...... (1) CQ = r + 4 √[(a - 10)2 + (b - 0)2] = r + 4 r = -4 + √(a2 + b2 - 20a + 100) ...... (2) CP = r √[(a - 1)2 + (b + 14)2] = r r = √(a2 + b2 - 2a + 28b + 197) ...... (3) (1) = (2) -1 + √(a2 + b2) = -4 + √(a2 + b2 - 20a + 100) 3 + √(a2 + b2) = √(a2 + b2 - 20a + 100) 9 + a2 + b2 + 6√(a2 + b2) = a2 + b2 - 20a + 100 6√(a2 + b2) = -20a + 91 ...... (4) (1) = (3) -1 + √(a2 + b2) = √(a2 + b2 - 2a + 28b + 197) [-1 + √(a2 + b2)]2 = [√(a2 + b2 - 2a + 28b + 197)]2 1 + a2 + b2 - 2√(a2 + b2) = a2 + b2 - 2a + 28b + 197 -2√(a2 + b2) = -2a + 28b + 196 6√(a2 + b2) = 6a - 84b - 588 ...... (5) (4) = (5) - 20a + 91 = 6a - 84b - 588 84b = 26a - 679 b = (26a - 679)/84 ...... (6) Subst. (6) into (4) 6√{a2 + [(26a - 679)/84]2} = -20a + 91 36{a2 + (26a - 679)2/7056} = (-20a + 91)2 36[7056a2 + (26a - 679)2] = 7056 (-20a + 91)2 36[7056a2 + (26a - 679)2] = 7056 (-20a + 91)2 [7056a2 + (26a - 679)2] = 196(-20a + 91)2 7056a2 + 676a2 - 35308a + 461041 = 196(400a2 - 3640a + 8281) 7056a2 + 676a2 - 35308a + 461041 = 78400a2 - 713440a + 1623076 70668a2 - 678132a + 1162035 = 0 a = [678132 √(6781322 - 4 x 70668 x 1162035)] / (2 x 70668) Hence, a = 7.362 ororor a = 2.233 When a = 7.362, subst. into (6) b = (26 x 7.362 - 679)/84 b = -5.805 (Rejected because the circle touches internally the circle with centre Q.) When a = 2.233, subst. into (6) b = (26 x 2.233 - 679)/84 b = -7.392 Ans: The centre of the required circle is (2.233, -7.392) =
其他解答:960E6207E048F01D
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