Question #5880

The line L is tangent to a circle, center at (2.5), at point (-1,1). Suppose line L is passing through (-5,4). Determine the equation of the line L.

Expert's answer

The line L is tangent to a circle, center at (2.5), at point (-1,1). Suppose line L is passing through (-5,4). Determine the equation of the line L.

If the required line L passes through the points (-1,1) and (-5,4), then it's equation is

(x-(-5)) / (-1-(-5)) = (y-4) / (1-4)

(x+5)/4 = (y-4)/(-3)

-3(x+5) = 4(y-4)

-3x - 15 = 4y - 16

4y + 3x = 1.

Let's check whether this line is tangent to a circle center at (2,5). If so, it will be normal to the line that passes through (2,5) and (-1,1):

(x-(-1)) / (2-(-1)) = (y-1) / (5-1)

(x+1)/3 = (y-1)/4

4(x+1) = 3(y-1)

4x + 4 = 3y - 3

3y - 4x = 7.

Here is the the scalar product of the direction vectors:

(4,3)*(3,-4) = 4*3 + 3*(-4) = 0.

We see that it is tangent to a circle. So, required line equation is 4y + 3x = 1.

If the required line L passes through the points (-1,1) and (-5,4), then it's equation is

(x-(-5)) / (-1-(-5)) = (y-4) / (1-4)

(x+5)/4 = (y-4)/(-3)

-3(x+5) = 4(y-4)

-3x - 15 = 4y - 16

4y + 3x = 1.

Let's check whether this line is tangent to a circle center at (2,5). If so, it will be normal to the line that passes through (2,5) and (-1,1):

(x-(-1)) / (2-(-1)) = (y-1) / (5-1)

(x+1)/3 = (y-1)/4

4(x+1) = 3(y-1)

4x + 4 = 3y - 3

3y - 4x = 7.

Here is the the scalar product of the direction vectors:

(4,3)*(3,-4) = 4*3 + 3*(-4) = 0.

We see that it is tangent to a circle. So, required line equation is 4y + 3x = 1.

## Comments

## Leave a comment