What is the equation of a circle of radius 1, which is also tangent to the line 3x-4y+1=0 at the y-axis value = 1 ?

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Answer 1 · 2015-06-19T21:14:02

There are two possible circles, either side of the line:

#(x-2/5)^2 + (y-9/5)^2 - 1 = 0#

#(x-8/5)^2 + (y - 1/5)^2 - 1 = 0#

When #y=1# the equation becomes #3x-4+1 = 0#, hence #x=1#.

So the circle is tangential to the line at #(1, 1)#.

Starting with #3x-4y+1 = 0#, add #4y# to both sides to get:

#4y = 3x+1#

Divide both sides by #4# to get:

#y = 3/4x+1/4#

The slope of this line is #m = 3/4#, so the slope of the radius of the tangential circle passing through the point of contact will be #-1/m = -4/3#.

So this perpendicular line will have equation:

#y = -4/3x + 7/3# in order that it passes through #(1,1)#

The centre of the tangential circle lies at distance #1# from #(1,1)# along this line.

That give two possibilities:

#(1 - 3/5, 1 + 4/5) = (2/5, 9/5)#

and

#(1 + 3/5, 1 - 4/5) = (8/5, 1/5)#

...using the right angled property of the 3, 4, 5 triangle.

The corresponding circles have equations:

#(x-2/5)^2 + (y-9/5)^2 - 1 = 0#

#(x-8/5)^2 + (y - 1/5)^2 - 1 = 0#

graph{(3x-4y+1)((x-2/5)^2 + (y-9/5)^2 - 1)((x-8/5)^2 + (y - 1/5)^2 - 1) = 0 [-3.52, 6.48, -1.28, 3.72]}