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

[Monoxdifly]

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?

 

[Greg]

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.

 

[Monoxdifly]

Wait... A tangent shouldn't go through the circle's center, right?

 

[Greg]

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.

 

[Monoxdifly]

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?

 

[Greg]

I don't see any errors in your work.

 

[Greg]

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?

 

[Monoxdifly]

Yes, the typo in my very first step.