Find the equation of the tangent

[Physicsissuef]

Homework Statement

Find the equation of the tangent of the circular [itex]K:x^2 + y^2 - 2x + 4y=0[/itex], perpendicular to the line x-2y+9=0.

Homework Equations

[itex](x_1-p)(x-p)+(y_1-q)(y-q)=r^2[/itex], equation of K.

[itex](kp-q+n)^2=r^2(k^2+1)[/itex], condition for tangent and circular K

The Attempt at a Solution

I tried like this. From the equation [itex]K:x^2 + y^2 - 2x + 4y=0[/itex],
p=1 and q=-2 ,[itex]r^2=5[/itex]
also from x-2y+9=0, the coefficient k of the equation of the tangent should be k=-2.
So I have y=-2x+n, and -2x-y+n=0

[itex](-2*1+2+n)^2=5(4+1)[/itex]
from here I get [itex]n_1=-5[/itex], and [itex]n_2=5[/itex], so the equation should be:

y=-2x-5

y=-2x+5

But the problem is that it is not same with my textbook results. Any help?

 

[physixguru]

Solve the equation of the tangent in terms of k with the equation of the given line to obtain a third point .

 

[HallsofIvy]

Homework Statement

Find the equation of the tangent of the circular [itex]K:x^2 + y^2 - 2x + 4y=0[/itex], perpendicular to the line x-2y+9=0.

Homework Equations

[itex](x_1-p)(x-p)+(y_1-q)(y-q)=r^2[/itex], equation of K.

Is [itex]x_1[/itex] different from x?

[itex](kp-q+n)^2=r^2(k^2+1)[/itex], condition for tangent and circular K

The Attempt at a Solution

I tried like this. From the equation [itex]K:x^2 + y^2 - 2x + 4y=0[/itex],
p=1 and q=-2 ,[itex]r^2=5[/itex]
also from x-2y+9=0, the coefficient k of the equation of the tangent should be k=-2.
So I have y=-2x+n, and -2x-y+n=0

[itex](-2*1+2+n)^2=5(4+1)[/itex]
from here I get [itex]n_1=-5[/itex], and [itex]n_2=5[/itex], so the equation should be:

y=-2x-5

y=-2x+5

But the problem is that it is not same with my textbook results. Any help?

Wouldn't it be simpler just to use basic concepts rather than such complicated formulas?

The line x- 2y+ 9= 0 can be written 2y= x+ 9 or y= (1/2)x+ 9/2 so its slope is 1/2. Any line perpendicular to that must have slope -2. For the given circle, [itex]x^2 + y^2 - 2x + 4y=0[/itex], implicit differentiation gives 2x+ 2yy'- 2+ 4y'= 0 or (2y+ 4)y'= 2- 2x so y'= (2- 2x)/(2y+ 4)= (1- x)/(y+ 2). For what values of x and y is that equal to 2?

You have two equations: (1-x)/(y+2)= 2 and [itex]x^2 + y^2 - 2x + 4y=0[/itex] to solve. Solve the first for either x or y, put that into the second equation and solve the resulting quadratic.

 

[Physicsissuef]

x=3
y=-3

x=1
y=-1

What to do now? x and y are different from x_1 and y_1