Question

So I was trying to find a general formula of a tangent line in a point "P" that belongs to a circumference.

Answer 1

This is what I have so far:

I began by taking in consideration the line that goes through point "o" and point "P".
so:
\[m _{op}=\frac{ \beta - y _{p} }{ \alpha - x _{p} }\]
And inserting it to a line:
\[(y-y _{p})=(\frac{ \beta - y _{p}{} }{ \alpha - x _{p} })(x-x _{p})\]

Answer 2

I know it's just the line that goes through the origin and point "P", I have to find the equation that belongs to the tangent line.

Answer 3

tangent is perpendicular to radius
did you know this ?

Answer 4

Yes, I was thinking of that.

Answer 5

since we know slope of OP, can you find the slope of tangent ?

Answer 6

so that means the slope of the tangent line is reciprocal and opposite sign han the line of the radius.

Answer 7

*product of slopes of perpendicular lines = -1*
and yes :)

Answer 8

so, if the slope is :
\[m _{op}=\frac{ \beta - y _{p} }{ \alpha - x _{p} }\]
Then the slope of that tangent line must be (I'll call it "l"):
\[m _{l}=-\frac{ \alpha - x _{p} }{ \beta - y _{p} }\]

Answer 9

correct.
you know slope, you have point, i am sure, you'll be able to find the equation of tangent line :)

Answer 10

Yes but I think I'll have to analyze it's intersection with the circumference.

Answer 11

point on tangent is xp,yp

y-yp=mi (x-xp)

Answer 12

mi or ml, whatever your slope was...

Answer 13

yes, like this:
\[y-y _{p}=-\frac{ \alpha - x _{p} }{ \beta - y _{p} }(x-x _{p})\]
So this must be valid:
\[(\beta- y_{p})(y-y_p)=-(x-x_p)(\alpha-x_p)\]
And also this:
\[(\beta - y_p)(y-y_p)+(x-x_p)(\alpha-x_p)=0\]
So applying distibutive and adding up things:
\[yy_p - y \beta - yy_p ^{2}+y_p \beta + xx_p - xp ^{2}- \alpha x + x_p \alpha =0\]
And the general equation for that circumference must be:
\[x ^{2}_p + y_p ^{2}-2 \alpha x_p - 2 \beta y_p +F = 0\]
So I would have to solve the system of equations:
\[yy_p - y \beta - yy_p ^{2}+y_p \beta + xx_p - xp ^{2}- \alpha x + x_p \alpha =0\]
\[x ^{2}_p + y_p ^{2}-2 \alpha x_p - 2 \beta y_p +F = 0\]

Answer 14

Adding both equalities I should get:
\[xx_p + yy_p - \alpha x - \alpha x_p - \beta y - \beta y_p + F=0\]
If I take common factor - alpha and - beta , divide and multiply by 2 I get:
\[xx_p + yy_p - 2 \alpha (\frac{ x+x_p }{ 2 })-2 \beta(\frac{ y+yp }{ 2 })+F=0\]
And that should represent the equation of any tangent line in any point "p" that belongs to the cicumference.
It looks the same as my teacher did, so yeah.
Thank you for helping me, I didin't take in mind that the line was perpendicular to the radius.

Answer 15

nice efforts! :)