The opposite slope means the opposite fraction, or opposite sign?

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Femme_physics
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Homework Statement



http://img64.imageshack.us/img64/7782/circleb.jpg

I'm supposed to find the slope equation of the tangent line to the circle at point A.
A = (2, 7)

The circle's formula is
(x+1)2+(y+3)2 = 25

The center point of the circle is therefor (-1, 3)

So the slope from the center point to A is

M = [tex]\frac{7-3}{2-(-1)}[/tex]
M = [tex]\frac{4}{3}[/tex]

That's true so far according to the answers.

So I want to plug [tex]\frac{4}{3}[/tex] for slope, just take the opposite sign of it, and from some reason in the solution they flipped 3 and 4 in the fraction for the slope of tangent line to the circle at point A! Here ->

http://img4.imageshack.us/img4/3995/answerofficial.jpg

That's wrong, am I right?
 
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No. You need to invert them. For example, think about the slope going from the center of the circle to the very top of the circle. Since you're going straight up, the slope must be infinite right (you go [tex]\Delta y = 5[/tex] but [tex]\Delta x =0[/tex])? Now, the tangent line at the top of the circle must have a slope of 0 since it's at the top of the circle right? So, the relationship between the two must be inverses. Making it the negative of it would make no sense because the tangent line's slope is obviously not [tex]-\infty[/tex]
 
Pengwuino said:
No. You need to invert them. For example, think about the slope going from the center of the circle to the very top of the circle. Since you're going straight up, the slope must be infinite right (you go [tex]\Delta y = 5[/tex] but [tex]\Delta x =0[/tex])? Now, the tangent line at the top of the circle must have a slope of 0 since it's at the top of the circle right? So, the relationship between the two must be inverses. Making it the negative of it would make no sense because the tangent line's slope is obviously not [tex]-\infty[/tex]

Brilliant explanation :)) thank you!
 
Femme_physics said:

Homework Statement



http://img64.imageshack.us/img64/7782/circleb.jpg

I'm supposed to find the slope equation of the tangent line to the circle at point A.
A = (2, 7)

The circle's formula is
(x+1)2+(y+3)2 = 25

The center point of the circle is therefor (-1, 3)
No, the center is at (-1, -3).
Femme_physics said:
So the slope from the center point to A is

M = [tex]\frac{7-3}{2-(-1)}[/tex]



M = [tex]\frac{4}{3}[/tex]

That's true so far according to the answers.

So I want to plug [tex]\frac{4}{3}[/tex] for slope, just take the opposite sign of it, and from some reason in the solution they flipped 3 and 4 in the fraction for the slope of tangent line to the circle at point A! Here ->

http://img4.imageshack.us/img4/3995/answerofficial.jpg

That's wrong, am I right?
The last image is confusing. What is the symbol after 8 1/2?
Also, mixed fractions such as 8 1/2 are seldom used in math texts because they could be interpreted as 8 + 1/2 or 8 * 1/2. Usually you see these as 15/2 or 7.5, but not as a mixed fraction.
 
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Femme_physics said:

Homework Statement



http://img64.imageshack.us/img64/7782/circleb.jpg

I'm supposed to find the slope equation of the tangent line to the circle at point A.
A = (2, 7)

The circle's formula is
(x+1)2+(y+3)2 = 25

The center point of the circle is therefor (-1, 3)

Mark44 said:
No, the center is at (-1, -3).

I suspect the OP made a typo in the equation for the circle. According to the diagram, the center M looks to be at (-1, 3). So the equation for the circle should be
[tex](x + 1)^2 + (y - 3)^2 = 25[/tex]
 
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