標題:
A.Math (Circles)
A straight line L with slope m, passes through a point P(-2,-2) and cuts a circle C at two points Q and R. The equation of the circle C is x^2+y^2-14x-10y+49=0 and the length of QR is 8 units.(a) Find, in terms of m, the equation of the line L.(b) By considering the shortest distance from the centre of the circle... 顯示更多 A straight line L with slope m, passes through a point P(-2,-2) and cuts a circle C at two points Q and R. The equation of the circle C is x^2+y^2-14x-10y+49=0 and the length of QR is 8 units. (a) Find, in terms of m, the equation of the line L. (b) By considering the shortest distance from the centre of the circle C to L, find the possible value(s) of m. Ans: (a) mx-y+(2m-2) = 0 (b) 4/3, 5/12
最佳解答:
a. Slope of the line L = m L passes through point P (-2 , -2) ∴ Equation of L: y - (-2) = m[x - (-2)] y + 2 = mx + 2m mx - y + (2m - 2) = 0 b. Centre of C: (7 , 5) Radius of C, r: 1/2 X √[(-14)2 + (-10)2 - 4(49)] = 5 Shortest distance between the centre of C and the line L, d = │[m(7) - (5) + (2m - 2)]/√(m2 + 12)│ = │(9m - 7)/√(m2 + 1)│ By Pythagoras theorem, d2 + (8/2)2 = r2 │(9m - 7)/√(m2 + 1)│2 + 16 = 25 (9m - 7)2/(m2 + 1) = 9 81m2 - 126m + 49 = 9m2 + 9 72m2 - 126m + 40 = 0 36m2 - 63m + 20 = 0 (3m - 4)(12m - 5) = 0 m = 4/3 or 5/12
其他解答:
A.Math (Circles)
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發問:A straight line L with slope m, passes through a point P(-2,-2) and cuts a circle C at two points Q and R. The equation of the circle C is x^2+y^2-14x-10y+49=0 and the length of QR is 8 units.(a) Find, in terms of m, the equation of the line L.(b) By considering the shortest distance from the centre of the circle... 顯示更多 A straight line L with slope m, passes through a point P(-2,-2) and cuts a circle C at two points Q and R. The equation of the circle C is x^2+y^2-14x-10y+49=0 and the length of QR is 8 units. (a) Find, in terms of m, the equation of the line L. (b) By considering the shortest distance from the centre of the circle C to L, find the possible value(s) of m. Ans: (a) mx-y+(2m-2) = 0 (b) 4/3, 5/12
最佳解答:
a. Slope of the line L = m L passes through point P (-2 , -2) ∴ Equation of L: y - (-2) = m[x - (-2)] y + 2 = mx + 2m mx - y + (2m - 2) = 0 b. Centre of C: (7 , 5) Radius of C, r: 1/2 X √[(-14)2 + (-10)2 - 4(49)] = 5 Shortest distance between the centre of C and the line L, d = │[m(7) - (5) + (2m - 2)]/√(m2 + 12)│ = │(9m - 7)/√(m2 + 1)│ By Pythagoras theorem, d2 + (8/2)2 = r2 │(9m - 7)/√(m2 + 1)│2 + 16 = 25 (9m - 7)2/(m2 + 1) = 9 81m2 - 126m + 49 = 9m2 + 9 72m2 - 126m + 40 = 0 36m2 - 63m + 20 = 0 (3m - 4)(12m - 5) = 0 m = 4/3 or 5/12
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