SOLUTION: Find the equation of a line tangent to the circle x^2+(y−3)^2=34 at the point (5, 0).
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Question 1126134: Find the equation of a line tangent to the circle x^2+(y−3)^2=34 at the point (5, 0).
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the equation of a line tangent to the circle x^2+(y-3)^2=34 at the point (5,0).
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x^2 + (y-3)^2=34
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2xdx + 2(y-3)dy = 0
dy/dx = -x/(y-3) --> the slope at any point on the circle.
@ (5,0), slope m = 5/3
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y = (5/3)*(x-5) is the tangent line.
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Without finding the derivative:
You can find the slope of the line from the center to the (5,0).
The slope of the tangent line is the negative inverse.
etc.
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