Question 1013853
Find the equation of the circle if the circle is tangent to the line 3x+y+2=0 at A(-1,1) and it passes through the point B(3,5).
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The center of the circle will be on the line perpendicular to the given line thru (-1,1).
y-1 = (1/3)*(x+1) is the line.
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The center is equidistant from A & B so it's on the perpendicular bisector (PB) of line AB.
The midpoint of AB is (1,3).
The slope of AB is 1.
--> PB is y-3 = -1*(x-1)
Or y = -x + 4
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The center is the intersection of y = -x + 4 and y-1 = (1/3)*(x+1)
-x + 4 = x/3 + 4/3
-3x + 12 = x + 4
x = 2
y = 2
--> the center is (2,2)
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The radius is the distance to A or B:
{{{r^2 = 3^2 + 1^1 = 10}}}
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{{{(x-2)^2 + (y-2)^2 = 10}}} is the circle.