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locus maths 3 Q
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1)the equation of two circles are given by C1:x^2 +y^2 +2x -8y=0C2: 2x^2 +2y^2 -12x -8y +18=0show that two circles intersect each other.2)a circles with centre O passes through three oints A(-5,0) B(19,0)andC(12,17)a)find the equation of the circle in the general formb)hence find the coordinates of... 顯示更多 1)the equation of two circles are given by C1:x^2 +y^2 +2x -8y=0 C2: 2x^2 +2y^2 -12x -8y +18=0 show that two circles intersect each other. 2)a circles with centre O passes through three oints A(-5,0) B(19,0)andC(12,17) a)find the equation of the circle in the general form b)hence find the coordinates of O and the radius of the circles . 3) a straight line L:kx +2y -5=0 and a circle (x-2)^2+(y-6)^2=1 ,P(1,1) is a point on L if Q is moving piont on L,which is the minimum distance between Q and the circle ?
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locus maths 3 Q
發問:
1)the equation of two circles are given by C1:x^2 +y^2 +2x -8y=0C2: 2x^2 +2y^2 -12x -8y +18=0show that two circles intersect each other.2)a circles with centre O passes through three oints A(-5,0) B(19,0)andC(12,17)a)find the equation of the circle in the general formb)hence find the coordinates of... 顯示更多 1)the equation of two circles are given by C1:x^2 +y^2 +2x -8y=0 C2: 2x^2 +2y^2 -12x -8y +18=0 show that two circles intersect each other. 2)a circles with centre O passes through three oints A(-5,0) B(19,0)andC(12,17) a)find the equation of the circle in the general form b)hence find the coordinates of O and the radius of the circles . 3) a straight line L:kx +2y -5=0 and a circle (x-2)^2+(y-6)^2=1 ,P(1,1) is a point on L if Q is moving piont on L,which is the minimum distance between Q and the circle ?
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(1) Centre of C1 = (-1, 4) Centre of C2 = (6, 4) Distance between centres = √[(-1 - 6)2 + (4 - 4)2] = 7 Radius of C1 = (1/2)√[22 + (-8)2] = √17 Radius of C2 = (1/2)√[(-12)2 + (-8)2 - 4(18)] = √34 Sum of radius = √17 (√2 + 1) > 7 Difference of radius = √17 (√2 - 1) < 7 So the circles intersect each other. (2a) Let the equation be x2 + y2 + Dx + Ey + F = 0 Then sub points A, B and C intro it individually. A: 25 - 5D + F = 0 ... (1) B: 361 + 19D + F = 0 ... (2) C: 433 + 12D + 17E + F = 0 ... (3) Solving, we have: D = 14, E = -38 and F = 45 So the equation is x2 + y2 + 14x - 38y + 45 = 0 (b) O is at (-7, 19) and then radius is: R = (1/2)√[142 + (-38)2 - 4(45)] = √365 (3) Sub P into L: k + 2 - 5 = 0 k = 3 So L is 3x + 2y - 5 = 0 Centre of circle = (2, 6) and radius = 1 So perpendicular distance from the centre to L is: |3(2) + 2(6) - 5|/√(32 + 22) = √13 Then the minimum distance between Q and the circle is √13 - 1.其他解答:
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