Question 1078578: What is the slope for the line tangent to the circle x^2 + y^2 = 8 at the point (2, -2)?
Found 3 solutions by ikleyn, Boreal, josmiceli:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
The radius-vector from the center to the point (2,-2) has the slope  = -1.
The tangent line to the circle at the tangent point is perpendicular to the radius.
Hence, the tangent line has the slope 1, reciprocal and opposite by the sign to the slope of the radius.
Answer. 1.
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! formula for circle at this point is y=-sqrt (8-x^2)
The derivative is -2x/-2 sqrt(8-x^2); when x=2, derivative equals 2/sqrt(4)=1, so the slope is 1
Answer by josmiceli(19441)  (Show Source):
You can put this solution on YOUR website! First, I'll check to see if (2,-2) is a point
on the circle
So, (2,-2) is a point on the circle
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Now I have to find the center of the circle
The general form for a circle is:
where the center is at (h,k)
I can rewrite the given equation as:
So, the center of the circle is at (0,0)
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Now I need to find the slope of the radius line
from (0,0) to (2,-2)
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The line tangent to the circle at (2,-2) will have slope=
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check the math, and get a 2nd opinion if needed
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