# (4,1) is one end of diametre of a circle and the tangent through the point at the other end of the diameter has equation 3x-y=1. Determine the...(4,1) is one end of diametre of a circle and the...

(4,1) is one end of diametre of a circle and the tangent through the point at the other end of the diameter has equation 3x-y=1. Determine the...

(4,1) is one end of diametre of a circle and the tangent through the point at the other end of the diameter has equation 3x-y=1. Determine the coordinates of the centre of the circle.

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Given one endpoint of a diameter of a circle is at (4,1), and the other end of that diameter is the point of tangency to the line 3x-y=1. Determine the coordinates of the center of the circle.

The tangent line has equation 3x-y=1 or y=3x-1.

The line that contains the diameter is perpendicular to the tangent line at the point of tangency.

The slope of the diameter will be `-1/3` as the slopes of perpendicular lines are opposite reciprocals of each other. (Their product is -1). The line containing the diameter goes through the point (4,1). So the equation of theline containing the diameter is found using the point-slope form:

`y-1=-1/3(x-4)==>y=-1/3x+4/3+3/3==>y=-1/3x+7/3`

Then the lines `y=3x-1` and `y=-1/3x+7/3` meet at the other end of the diameter, the point of tangency. Solving simultaneously we get:

`3x-1=-1/3x+7/3 ==>10/3x=10/3==>x=1` , and so y=2.

The center of thecircle is at the midpoint of the diameter, the segment whose endpoints are at (4,1) and (1,2). The midpoint of a segment is found at the average of the x's and y's, so the center is at `((4+1)/2,(2+1)/2)` or `(5/2,3/2)`

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**The center of the circle is at `(5/2,3/2)` **

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The graph: