Find an equation of the line?

Question

How do you find the equation of the line that is tangent to the circle x^2+y^2-4x-8y-5=0 at the point (-1,0)?

Answer 1

Easy, get the first derivative of the eqn, which would be :

2x + 2y.y' -4 -8y' = 0

then substitute with the given point to get the slope of the line :

2(-1) + 2(0).y' -4 -8y' =0

-2 - 4 -8y' =0

y ' = 3/4 (which is the slope (m) of the tangent at the given point)

since y' at (-1, 0) = m = (y-y1) / (x-x1)

therefore : 3/4 = (y-0) / (x +1)

cross multiplication yields :

3x + 3 = 4y ---> this is the eqn you are asking for.

Answer 2

There is a shortcut to this, but I forgot and hence I shall be giving you the longer answer.

The equation of the circle is
x^2 + y^2 -4x -8y -5 = 0
=> (x - 2)^2 + (y - 4)^2 - 4 - 16 - 5 = 0
=> (x - 2)^2 + (y-4)^2 = 25

Center of the circle is (2,4) and radius is 5 units.

To calculate equation of the line, we need slope and y-intercept.

y = mx + c

Tangent of the circle is perpendicular to the radius. Hence, the slope of the tangent is the negative inverse of the slope of the radial line from (-1,0) to (2,4)

Slope of radial line = (0 - 4) / (-1 - 2) = 4 / 3

Slope of the tangent = m = -3 / 4

To determine c, we know that the line passes through (-1,0). Substitute it in the equation of the line above as follows:

0 = (-3/4)(-1) + c
=> c = 3/4

Therefore, equation of the line is:

y = (-3/4)*x + 3/4

or

3x + 4y = 3


***********************************************************************
In the solution proposed by M T, there is a tiny error.

Easy, get the first derivative of the eqn, which would be :

2x + 2y.y' -4 -8y' = 0

then substitute with the given point to get the slope of the line :

2(-1) + 2(0).y' -4 -8y' =0

-2 - 4 -8y' =0

y ' = 3/4 (which is the slope (m) of the tangent at the given point)


y' should actually equal -3/4


The rest of the solution is fine.

Answer 3

For these questions, it is easier to draw a picture first.
All you're looking for is one vector (i.e. the direction) of your line. You know it is tangent to the circle, therefore it is perpendicular to the radius that goes to the point of tangence - write your vector as (1,y) and get y by expressing this condition.
Once you have that, the points of your line are the sets of points of the form (-1,0)+k(1,y) - it is easy to derive an equation from there.
I hope that helped

Answer 4

If you have a cf.with equation

x^2+y^2-2ax-2by+c= 0 The equ. af the tangent at a point(p,q)of it

is px+qy-a(x+p)-b(y+q)+c=0 in this case

-x -2(x-1)-4y-5 =0 -3x-4y-3=0 y= -3/4*x-3/4