What Is the Correct Equation for the Tangent of a Circle at a Given Point?

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Discussion Overview

The discussion revolves around finding the correct equation for the tangent line to a circle at a specified point. Participants explore the mathematical steps involved in deriving the tangent line equations, including the identification of the circle's center and the slopes of the tangent lines.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes to find the tangent line equation of the circle defined by $$x^2+y^2+6x-8y+12=0$$ at the point where the x-coordinate is -1.
  • Another participant suggests finding the center of the circle and the equations of lines passing through the center and the points (-1, -1) and (-1, -7).
  • A participant questions whether a tangent line should pass through the center of the circle, indicating a potential misunderstanding in the approach.
  • Further clarification is provided regarding the need to find the slopes of the lines connecting the center to the specified points and then determining the slopes of the tangent lines.
  • One participant calculates the center of the circle as (-3, 4) and derives slopes for the tangent lines, leading to equations that are then questioned for accuracy.
  • Another participant acknowledges the calculations but expresses confusion about a possible error in the earlier steps.
  • A later reply identifies a typo in the initial calculations as a source of error.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correct approach to finding the tangent line equations, with multiple interpretations and calculations presented without consensus on the final answer.

Contextual Notes

Participants' calculations depend on the correct identification of the circle's center and the slopes of the tangent lines, which remain unresolved due to the identified errors and differing interpretations of the problem.

Monoxdifly
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One of the tangent line equation of the circle $$x^2+y^2+6x-8y+12=0$$ at the point whose absis is -1 is ...
A. 2x - 3y - 7 = 0
B. 2x - 3y + 7 = 0
C. 2x + 3y - 5 = 0
D. 2x - 3y - 5 = 0
E. 2x - 3y + 5 = 0

By substituting x = -1, I got:
$$(-1)^2+y^2+6(-1)+8y+12=0$$
$$1+y^2-6+8y+12=0$$
$$y^2+8y+7=0$$
(y + 1) (y + 7) = 0
y = -1 or y = -7
Then what?
 
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Find the center $(h,k)$ of the circle and then find the equations of the two lines that pass through $(h,k),(-1,-1)$ and $(h,k),(-1,-7)$. One of those equations should match one of the answer choices.
 
Wait... A tangent shouldn't go through the circle's center, right?
 
That's right...my mistake. Find the slopes of the lines that pass through the points I gave. Call them $m$. Then find the equations of the lines that have slope $-\frac1m$ and pass through $(-1,-1)$ and $(-1,-7)$. That should get you your answer.
 
Let's try... Judging by the circle's equation, the center is (-3, 4) thus the slope you asked is either $$-\frac{5}{2}$$ or $$-\frac{11}{2}$$ and so the slope of the tangent is either $$\frac{2}{5}$$ or $$\frac{2}{11}$$. Trying to substitute the first one I'll get:
$$y-(-1)=\frac{2}{5}(x-(-1))$$
$$y+1=\frac{2}{5}(x+1)$$
5(y + 1) = 2(x + 1)
5y + 5 = 2x + 2
5y - 2x + 3 = 0
For the second one:
$$y-(-7)=\frac{2}{11}(x-(-1))$$
$$y+7=\frac{2}{11}(x+1)$$
11(y + 7) = 2(x + 1)
11y + 77 = 2x + 2
11y - 2x + 73 = 0
Huh? Where did I go wrong? Or did I misinterpret your hint?
 
I don't see any errors in your work.
 
Monoxdifly said:
One of the tangent line equation of the circle $$x^2+y^2+6x-8y+12=0$$ at the point whose absis is -1 is ...
A. 2x - 3y - 7 = 0
B. 2x - 3y + 7 = 0
C. 2x + 3y - 5 = 0
D. 2x - 3y - 5 = 0
E. 2x - 3y + 5 = 0

By substituting x = -1, I got:
$$(-1)^2+y^2+6(-1)+8y+12=0$$
$$1+y^2-6+8y+12=0$$
$$y^2+8y+7=0$$
(y + 1) (y + 7) = 0
y = -1 or y = -7
Then what?

There's an error in the above. Can you spot it?
 
Yes, the typo in my very first step.
 

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