SOLUTION: Given the line 𝑥 + 2𝑦 − 4 = 0 and circle 𝑥^2 + 𝑦^2 − 8𝑥 − 4 = 0, find the equations of the tangents to the circles which are perpendicular to the line. Sketch

The given circle equation is

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



By completing the square, you reduce it to the standard form circle equation

    (x-4)^2 + y^2 = 20.


So, the circle has the center at the point C= (4,0) and has the radius of   = 4.472 (rounded).

Notice that the circle intersects y-axis at the points y= +/-2. So, y-intercepts are the points A= (0,2) and B= (0,-2).



The given line x + 2y - 4 = 0 has the slope  m= -0.5.

Notice that this line goes through the center C= (4,0) of the circle and through the point A= (0,2).
so point A is actually an intersection point of the given line and the circle.



Thus we discovered very important fact: the given line goes through the center of the given circle.


    +-----------------------------------------------------------------------+
    |    It implies that the lines tangent to the circle and perpendicular  |
    |    to the given line factually go through the intersection points     |
    |               of the given line and the circle.                       |
    +-----------------------------------------------------------------------+



Thus the desired lines have the slope  =  = 2, since they are perpendicular to the given line.


Next, one intersection point of the given line and the circle is the point A= (0,2).
Hence, the other intersection point of the given line and the circle is D= (8,-2), 
symmetrical to A relative the center of the circle C= (4,0).


Thus first  desired line has the slope 2 and goes through A= (0,2).  Its equation is  y-2 = 2x,
     
The second  desired line has the slope 2 and goes through D= (8,-2). Its equation is  y+2 = 2(x-8).


ANSWER.  First  desired line has equation  y-2 = 2x.

         Second  desired line has equation  y+2 = 2(x-8).

         These equations can be presented in any other equivalent forms.