maths ,geometric relationship

[匿名]

2015-01-27 6:06 am

GIVEN THAT THE EQUATION OF CIRCLE C IS x^2+y^2-10x-22y+46=0
Denote the centre of C by Q
The equation of a straight line L is 12x+5y+54=0 , where L and C does not intersect.Let P be a point of L which is the closest to Q .
so the corridinate of P is (-7,6)
the question is :Let R be a moving point on C such that it is farthest from P

1)describe the geometric relationship between P,Q and R

2)find the ratio of the area of triangle OPQ and the triangle of OQR ,where O is the origin

回答 (1)

用戶10012

2015-01-27 3:38 pm

✔ 最佳答案

Co- ordinates of Q is ( 5, 11).
Since P ( - 7, 6) is closest to Q, PQ must be perpendicular to the straight line L.
(You can prove that slope of PQ x slope of L = - 1).
The farthest point from P and on the circle must be the point on PQ as well. Therefore PQR is a straight line with RQ equals to radius of circle.
Length of RQ = radius of circle = sqrt [ 25 + 121 - 46] = 10
Length of QP = sqrt [ (5 + 7)^2 + (11 - 6)^2] = sqrt ( 144 + 25) = sqrt 169 = 13.
So ratio of area of triangle OPQ to area of triangle OQR = 13 : 10.
[ Note : Ratio of area of 2 triangle with SAME HEIGHT = ratio of the their base length.]

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