SOLUTION: Find the equation of the circle if the circle is tangent to the line 3x+y+2=0 at (-1,1) and it passes through the point (3,5).

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the equation of the circle if the circle is tangent to the line 3x+y+2=0 at (-1,1) and it passes through the point (3,5).      Log On


   



Question 1013853: Find the equation of the circle if the circle is tangent to the line 3x+y+2=0 at (-1,1) and it passes through the point (3,5).
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
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%5E2+=+3%5E2+%2B+1%5E1+=+10
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%28x-2%29%5E2+%2B+%28y-2%29%5E2+=+10 is the circle.