# Find the equation of the tangent

1. Apr 12, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$(x_1-p)(x-p)+(y_1-q)(y-q)=r^2$, equation of K.

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

3. The attempt at a solution

I tried like this. From the equation $K:x^2 + y^2 - 2x + 4y=0$,
p=1 and q=-2 ,$r^2=5$
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

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

y=-2x-5

y=-2x+5

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

2. Apr 12, 2008

### physixguru

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

3. Apr 12, 2008

### HallsofIvy

Staff Emeritus
Is $x_1$ different from x?

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, $x^2 + y^2 - 2x + 4y=0$, 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 $x^2 + y^2 - 2x + 4y=0$ to solve. Solve the first for either x or y, put that into the second equation and solve the resulting quadratic.

4. Apr 12, 2008

### Physicsissuef

x=3
y=-3

x=1
y=-1

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