Question 1064393: Find the equation of the circle satisfying the given conditions (general equation)
1. Tangent to the line 4x-3y=6 at (3,2) and passing through (2,-1).
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the equation of the circle satisfying the given conditions (general equation)
1. Tangent to the line 4x-3y=6 at (3,2) and passing through (2,-1).
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Find the line thru (3,2) perpendicular to the given line.
The center will be on that line.
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Find the perpendicular bisector of the line between (3,2) and (2,-1).
The center is also on that line.
The intersection of the 2 lines is the center, (h,k).
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The distance from the center to either point is the radius r.
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The 2 lines are
x + 3y = 4 Eqn A
3x + 4y = 17 Eqn B
3x + 9y = 12 Eqn A times 3
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-5y = 5
y = -1
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x = 7
--> center @ (7,-1)
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Using the point (3,2):
r^2 = diffy^2 + diffx^2 = 9 + 16 = 25
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 is the circle.
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I'll check it tomorrow.
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