Given two circles C1:x^2+y^2+2x-6y+5=0 and C2:x^2+y^2-6x-2y+5=0 which touch each other at P.
(a) Find the coordinates of P.
(b) Find the equation of the common tangent at P.
equation of circle
2007-10-02 7:47 am
回答 (2)
2007-10-02 8:14 am
參考: My Maths knowledge
2007-10-02 8:08 am
(a) P is the point of the division of the line segment joining the centres with ratio r1 : r2 (r1 = radius of C1, r2 = radius of C2_
radius of C1 : 開方(1^2 + (-3)^2 - 5) = 開方(5)
radius of C2 : 開方((-3)^2 + (-1)^2 - 5) = 開方(5)
So P is the mid-point of the line segment joining the centres
P( (-1 + 3)/2, (3 +1)/2) = (1, 2)
(b) slope of line segment joining the centres
(1 - 3)/(3 - (-1)) = -1/2
slope of tangent = 2
Equation of tangent : y - 2 = 2(x -1)
2x - y + 1 = 0
radius of C1 : 開方(1^2 + (-3)^2 - 5) = 開方(5)
radius of C2 : 開方((-3)^2 + (-1)^2 - 5) = 開方(5)
So P is the mid-point of the line segment joining the centres
P( (-1 + 3)/2, (3 +1)/2) = (1, 2)
(b) slope of line segment joining the centres
(1 - 3)/(3 - (-1)) = -1/2
slope of tangent = 2
Equation of tangent : y - 2 = 2(x -1)
2x - y + 1 = 0
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