Question 194599: give the equation of a circle tangent to the line 3x+y-2=0 at (-1,5) and with radius of 3.16227766
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! give the equation of a circle tangent to the line 3x+y-2=0 at (-1,5) and with radius of 3.16227766
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There are 2 circles, one on either side of the line.
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3x+y-2=0 has a slope, m, of -3, so the line thru (-1,5) and the center has a slope of +1/3.
The center is a distance of sqrt(10) from (-1,5)
The eqn of the line thru the center and the point (-1,5) is found by:
y-y1 = m*(x-x1) where (x1,y1) is the point (-1,5)
y-5 = (1/3)*(x+1)
y = x/3 + 16/3
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The center is sqrt(10) units from (-1,5) along this line.
Using Pythagoras:
10 = (x+1)^2 + (y-5)^2
Sub for y
10 = (x+1)^2 + (((x+16)/3) - 5)^2)
10 = x^2 + 2x + 1 + (1/9)*(x^2 + 2x + 1)
90 = 9x^2 + 18x + 9 + x^2 + 2x + 1
90 = 10x^2 + 20x + 10
x^2 + 2x - 8 = 0
x = -4, x = 2
Centers at (-4,4) and at (2,6)
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Center at (-4,4): (x+4)^2 + (y-4)^2 = 10
Center at (2,6): (x-2)^2 + (y-6)^2 = 10
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