標題:
數學f5求解plz~
發問:
1) let a be a constant, The circle x^2+y^2-(a+2)x-5ay+19=0 passes throught P(-1,6). The equation of the tangent to the circle at the point P.2) The straight line 4x-3y-54=0 and the circle x^2+y^2-10x+16y-(2k+49)=0 have point of intersection, where k is a constant. Find the value of k.3) Consider two points... 顯示更多 1) let a be a constant, The circle x^2+y^2-(a+2)x-5ay+19=0 passes throught P(-1,6). The equation of the tangent to the circle at the point P. 2) The straight line 4x-3y-54=0 and the circle x^2+y^2-10x+16y-(2k+49)=0 have point of intersection, where k is a constant. Find the value of k. 3) Consider two points A(0,4) and B (3,-1) on a rectangular coordinate plane. A point P moves on the same plane such that AP:PB=1:2. Find the equation of the locus of P. 4) The equation of a circle is 3x^2+3y^2-12x+24y-87=0. Find the radius of the circle. 5) The line segment joining P(-12,-9) and Q(2,5) is a diameter of a circle. Find the equation of the circle.
1) Put (-1,6) into the equation of the circle, we have (-1)^2 + 6^2 - (a + 2)(-1) - 5a(6) + 19 = 0 a = 2 The equation of the circle becomes x^2 + y^2 - 4x - 10y + 19 = 0 (x - 2)^2 + (y - 5)^2 = 10 The coordinates of center O = (2, 5) Slope of OP = (6 - 5)/(-1 - 2) = -1/3 Slope of tangent at P = -1/(-1/3) = 3 The equation of tangent at P: y - 6 = 3(x + 1) y = 3x + 9 2) The straight line: 4x - 3y - 54 = 0 y = (4x - 54)/3 ... (i) The circle: x^2 + y^2 - 10x + 16y - (2k + 49) = 0 ... (ii) Put (i) into (ii): x^2 + (1/9)(4x - 54)^2 - 10x + (16/3)(4x - 54) - (2k + 49) = 0 (25/9)x^2 - (110/3)x - (2k + 13) = 0 Since there is/are point(s) of intersection, [-(110/3)]^2 - 4(25/9)[-(2k + 13)] ≥ 0 12100/9 + (100/9)(2k + 13) ≥ 0 121 + 2k + 13 ≥ 0 2k ≥ -134 k ≥ -67 Note: I'm not sure about the statement "have point of intersection". I think it is to tell there are "at least one point of intersection" rather than "have a point of intersection". If there is only one point of intersection, k = -67. 3) 2AP = PB Locus of P: 2[(x - 0)^2 + (y - 4)^2]^(1/2) = {(x - 3)^2 + [y - (-1)]^2}^(1/2) 4x^2 + 4(y - 4)^2 = (x - 3)^2 + (y + 1)^2 4x^2 + 4y^2 - 32y + 64 = x^2 - 6x + 9 + y^2 + 2y + 1 3x^2 + 3y^2 + 6x - 34y + 54 = 0 4) 3x^2 + 3y^2 - 12x + 24y - 87 = 0 x^2 + y^2 - 4x + 8y - 29 = 0 [(x - 2)^2 - 4] + [(y + 4)^2 - 16] - 29 = 0 (x - 2)^2 + (y + 4)^2 = 49 = 7^2 i.e. radius = 7. 5) Coordinates of center O = Coordinates of mid-point of PQ = [(-12 + 2)/2, (-9 + 5)/2] = (-5, -2) Length of radius = OQ = [(-5 - 2)^2 + (-2 - 5)^2]^(1/2) = [2(7^2)]^(1/2) = 7[2^(1/2)] i.e. The equation of the circle: [x - (-5)]^2 + [y - (-2)]^2 = {7[2^(1/2)]}^2 (x + 5)^2 + (y + 2)^2 = 98 Or you may express in general form: x^2 + y^2 + 10x + 4y - 69 = 0
其他解答:
數學f5求解plz~
發問:
1) let a be a constant, The circle x^2+y^2-(a+2)x-5ay+19=0 passes throught P(-1,6). The equation of the tangent to the circle at the point P.2) The straight line 4x-3y-54=0 and the circle x^2+y^2-10x+16y-(2k+49)=0 have point of intersection, where k is a constant. Find the value of k.3) Consider two points... 顯示更多 1) let a be a constant, The circle x^2+y^2-(a+2)x-5ay+19=0 passes throught P(-1,6). The equation of the tangent to the circle at the point P. 2) The straight line 4x-3y-54=0 and the circle x^2+y^2-10x+16y-(2k+49)=0 have point of intersection, where k is a constant. Find the value of k. 3) Consider two points A(0,4) and B (3,-1) on a rectangular coordinate plane. A point P moves on the same plane such that AP:PB=1:2. Find the equation of the locus of P. 4) The equation of a circle is 3x^2+3y^2-12x+24y-87=0. Find the radius of the circle. 5) The line segment joining P(-12,-9) and Q(2,5) is a diameter of a circle. Find the equation of the circle.
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最佳解答:1) Put (-1,6) into the equation of the circle, we have (-1)^2 + 6^2 - (a + 2)(-1) - 5a(6) + 19 = 0 a = 2 The equation of the circle becomes x^2 + y^2 - 4x - 10y + 19 = 0 (x - 2)^2 + (y - 5)^2 = 10 The coordinates of center O = (2, 5) Slope of OP = (6 - 5)/(-1 - 2) = -1/3 Slope of tangent at P = -1/(-1/3) = 3 The equation of tangent at P: y - 6 = 3(x + 1) y = 3x + 9 2) The straight line: 4x - 3y - 54 = 0 y = (4x - 54)/3 ... (i) The circle: x^2 + y^2 - 10x + 16y - (2k + 49) = 0 ... (ii) Put (i) into (ii): x^2 + (1/9)(4x - 54)^2 - 10x + (16/3)(4x - 54) - (2k + 49) = 0 (25/9)x^2 - (110/3)x - (2k + 13) = 0 Since there is/are point(s) of intersection, [-(110/3)]^2 - 4(25/9)[-(2k + 13)] ≥ 0 12100/9 + (100/9)(2k + 13) ≥ 0 121 + 2k + 13 ≥ 0 2k ≥ -134 k ≥ -67 Note: I'm not sure about the statement "have point of intersection". I think it is to tell there are "at least one point of intersection" rather than "have a point of intersection". If there is only one point of intersection, k = -67. 3) 2AP = PB Locus of P: 2[(x - 0)^2 + (y - 4)^2]^(1/2) = {(x - 3)^2 + [y - (-1)]^2}^(1/2) 4x^2 + 4(y - 4)^2 = (x - 3)^2 + (y + 1)^2 4x^2 + 4y^2 - 32y + 64 = x^2 - 6x + 9 + y^2 + 2y + 1 3x^2 + 3y^2 + 6x - 34y + 54 = 0 4) 3x^2 + 3y^2 - 12x + 24y - 87 = 0 x^2 + y^2 - 4x + 8y - 29 = 0 [(x - 2)^2 - 4] + [(y + 4)^2 - 16] - 29 = 0 (x - 2)^2 + (y + 4)^2 = 49 = 7^2 i.e. radius = 7. 5) Coordinates of center O = Coordinates of mid-point of PQ = [(-12 + 2)/2, (-9 + 5)/2] = (-5, -2) Length of radius = OQ = [(-5 - 2)^2 + (-2 - 5)^2]^(1/2) = [2(7^2)]^(1/2) = 7[2^(1/2)] i.e. The equation of the circle: [x - (-5)]^2 + [y - (-2)]^2 = {7[2^(1/2)]}^2 (x + 5)^2 + (y + 2)^2 = 98 Or you may express in general form: x^2 + y^2 + 10x + 4y - 69 = 0
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