Equation of Circles

用戶225241

2008-06-23 9:08 pm

1) Find the equation of a circle with centre at(-2,3) and tangent to the line
2x-y+3 = 0. Hence find the coordinates of the point at which the circle
touches the tangent.

2) Find the equation of the circle through the intersection of the circle
x^2 + y^2 -2x -4y -12 = 0 and the line x-3y+2 = 0 with radius 2開方2

回答 (1)

[匿名]

2008-06-23 10:08 pm

✔ 最佳答案

(1) r^2=[(2*-2-3+3)/(開方(4+1)]
=16/5

(x+2)^2+(y-3)^2=16/5
x^2 + y^2+4x-6y+4+9-16/5=0
x^2 + y^2+4x-6y+49/5

then find eq. of normal : (y-3)/(x+2)=-1/2 ( slope of tangent=2)
x+2y-4=0
solve {x+2y-4=0
2x-y+3 = 0
we have coordinates of the point =(-2/5, 11/5)

(2) By family of cirlce
x^2 + y^2 -2x -4y +1+k(x-3y+2 )=0.......(i)
x^2 + y^2-x(2-k)-y(4+3k)+1+2k=0
r^2=[(2-k)/2]^2+[(4+3k)/2}^2-(1+2k)
8 =(10k^2+12k+16)/4
5k^2+6k-8=0
k= -2 or 4/5
sub k=-2 into (i)
we have x^2+y^2-4x+2y-3=0
similary sub. k=4/5 int (i)
we have x^2+y^2 -(6/5)x-(32/5)y+13/5=0

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