SOLUTION: Find the equation of a circle passing through (-1,6) and tangent to the lines x-2y+8=0 and 2x+y+6=0.

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Question 1015843: Find the equation of a circle passing through (-1,6) and tangent to the lines x-2y+8=0 and 2x+y+6=0.
Found 2 solutions by FrankM, Alan3354:
Answer by FrankM(1040) (Show Source):
You can put this solution on YOUR website!
There is no solution. Please see this graph.

The red line is the equation x-2y+8=0. The point (-1,6) forms a perpendicular line to it at (0,4) which is away.
The circle equation is but as you see, it's tangent to the one line, not both at once.

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
The other tutor misread it, and used (-1,6) as the center of the circle.
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Find the equation of a circle passing through (-1,6) and tangent to the lines x-2y+8=0 and 2x+y+6=0
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Find the intersection of the 2 lines.
x - 2y + 8 = 0
2x + y + 6 = 0
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x - 2y + 8 = 0
4x +2y + 12 = 0 2nd eqn times 2
----------------------------------- Add
5x + 20 = 0
x = -4
y = 2
--> (-4,2)
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The center of the circle will be on the bisector of the 2 lines, the point (h,k).
The slopes of the lines are 22 and 1/2 --> they're perpendicular.
The angle of x - 2y + 8 = 0 with the x-axis is arctan(1/2) =~ 26.565 degs
Add 45 --> 71.565 degs
m = tan(71.565) = 3
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Find the eqn of the bisector: m = 3 thru (-4,2)
--> y-2 = 3(x+4)
The center of the circle is on this line, and is equidistant from the 2 given lines and the point (-1,6)
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Using the line x - 2y + 8 = 0
Distance to the line d = |h - 2k + 8|/sqrt(5)
Distance to (-1,6) = sqrt((h+1)^2 + (k-6)^2)
sqrt((h+1)^2 + (k-6)^2) = |h - 2k + 8|/sqrt(5)
Square both sides
(h+1)^2 + (k-6)^2 = |h - 2k + 8|^2/5
h^2 + 2h + 1 + k^2 - 12k + 36 = (h^2 - 4hk + 6k^2 - 16h - 32k + 64)/5
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That's a mess.
I'll work on this later.