SOLUTION: Hi Quick question if somebody would be so kind as to walk me through it Find the equation of the tangent to the circle x^2+y^2-4y-1=0 The the point (2,1) Thanks i

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Question 358061: Hi
Quick question if somebody would be so kind as to walk me through it
Find the equation of the tangent to the circle
x^2+y^2-4y-1=0
The the point (2,1)
Thanks in advance
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Find the equation of the tangent to the circle
x^2+y^2-4y-1=0
The the point (2,1)
------------------
Find the derivative.
2x + 2yy' - 4y' = 0
y'(2y-4) = -2x
---
y' = -2x/(2y-4)
This is the slope at every point (x,y)
--------------------------------------------
Let x = 2 ; Let y = 1 to find the slope at (2,1)
y' = -4/(-2) = 2
----
Find the line with slope = 2 passing thru (2,1)
1 = -2(2)+b
b = 5
-------
Equation of the tangent at (2,1)
y = 2x+5
========================
Cheers,
Stan H.
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Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

In the first place, what makes you think this is a "Quick Question"?

Since there is a term, but no term, you only need to complete the square on .

Add the opposite of the constant term to both sides:

Divide the coefficient on by 2, square the result, and add that result to both sides: (-4 divided by 2 is -2, -2 squared is 4]

Factor the perfect square in :

Rewrite:

Hence the circle has a center at and a radius

The tangent to a circle at a given point is perpendicular to the line containing the radius through the given point.

Find the slope of the line containing the radius by using the slope formula on the given point and the coordinates of the center of the circle:

where are the coordinates of the given point and are the coordinates of the circle center.

Since

calculate the negative reciprocal of the slope you calculated above.

Finally use the point-slope form of an equation of a straight line:

where are the coordinates of the given point and is calculated negative reciprocal of the radius slope.

Remember to put the result in the appropriate final form as specified by your instructor or text (if any).

John

My calculator said it, I believe it, that settles it