SOLUTION: Find the radius of the circle touching the straight lines x-2y-1 = 0 and 3x-6y+7 = 0. Please help. Thanks

Question 200697: Find the radius of the circle touching the straight lines x-2y-1 = 0 and
3x-6y+7 = 0.
Please help.
Thanks
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the radius of the circle touching the straight lines x-2y-1 = 0 and
3x-6y+7 = 0.
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The 2 lines are parallel. The slope, m, is 1/2. The slope is the tangent of the angle that the lines make with the x-axis.
Put the eqns into slope-intercept form, which means solve for y.
x-2y-1 = 0 --> y = (1/2)x - 1/2
3x-6y+7 = 0 --> y = (1/2)x + 7/6
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The difference in the y-intercepts is 1/2 + 7/6 = 5/3.
The distance between the 2 lines is 5/3 times the cosine of the angle.
d = (5/3)*cos(arctan(1/2))
d = ~ 5/3*0.89443 which is the diameter.
r = ~ 0.745
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There are an infinite number of circles that can be between these 2 lines.

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