SOLUTION: Dear Sir/Madam, I am confronted with the following problem: "Find the equation of the line that is tangent to the circle x^2 + y^2 = 25 at the point P(-3,4)." I did the

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Dear Sir/Madam, I am confronted with the following problem: "Find the equation of the line that is tangent to the circle x^2 + y^2 = 25 at the point P(-3,4)." I did the      Log On

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Question 6647: Dear Sir/Madam,
I am confronted with the following problem:
"Find the equation of the line that is tangent to the circle x^2 + y^2 = 25 at the point P(-3,4)."
I did the following:
1) +y%5E2+=+25+-+x%5E2+
2) +y+=+sqrt%2825+-+x%5E2%29+
3) f'(x) = +-2x%2F%282sqrt%2825+-+x%5E2%29%29+
4) f'(-3) = +-6%2F%282sqrt%2816%29%29+=+-6%2F8+=+-3%2F4+
5) +-3%2F4+=+%28y+-+4%29%2F%28x+%2B+3%29+
6) -3(x + 3) = 4(y - 4)
7) -3x - 9 = 4y - 16
8) 4y = -3x + 7
9) +y+=+-3x%2F4+%2B+7%2F4+ which should be the equation of the tangent.
Instead however, +y+=+3x%2F4+%2B+25%2F4+ is the equation of the tangent. Why?
Thanks in advance.
Regards,
-Mike
Found 2 solutions by rapaljer, Mike:
Answer by rapaljer(4551) About Me  (Show Source):
You can put this solution on YOUR website!
Your problem is in step 2, where you solved for y by taking the square root of both sides of the equation. You must include a "+ or -" symbol in this step, where the plus or the minus is determined by the point that is selected. What you have is the equation of a circle, and the point of tangency is at (-3,4) placing the point in the second quadrant, since x is negative and y is positive. Notice that in the second quadrant, the slope of a tangent line to a point on the curve will have a positive slope (by inspection!), so you have to use the plus sign for the slope of the tangent line. In quadrant IV, where x is positive and y is negative, you also have a positive slope. In quadrants I and III, the tangent line to a circle will have a negative slope (again by inspection!).

I think the rest of what you have done is correct.

R^2 at SCC

Answer by Mike(39) About Me  (Show Source):
You can put this solution on YOUR website!
The problem actually lies in step 4...-3 was plugged in wrongly. It should be 6 and not -6.