You can put this solution on YOUR website! Find an equation(s) of the line(s) containing (5,1) and at a distance 1 from the origin.
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--> 2 lines thru the point and tangent to the circle
One line is parallel to the x-axis
y = 1 *** eqn of one of the lines
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The slope of the line from (5,1) to the Origin (center of the 1st circle) = 1/5
= tangent of the angle between the line and the x-axis.
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The line from (0,0) to (5,1) is the bisector of the 2 tangent lines thru (5,1).
--> the angle of the 2nd line and the x-axis = 2x the bisector
angle = 2*atan(1/5)
slope = tan(2*atan(1/5)) = 5/12
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Y = mx + b
Use the point (5,1) and the slope = 5/12
1 = (5/12)*5 + b
b = -13/12
2nd eqn y = (5/12)x - 13/12
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y = 1 & y = 5x/12 - 13/12
