Question #f69b8

To show that a particular point is on the circle, simply substitute the coordinate values into the equation and see if the equation holds.

#x^2 + y^2 = 20#

#p = (2, -4)#

#=> 2^2 + (-4)^2 = 20#?

#=> 4 + 16 = 20#?

#=> 20 = 20#

A line tangent to the circle at point #p# is perpendicular to the line connecting the point #p# and the center of the circle.

To find the equation of the tangent at point #p#, first we get the slope of the line connecting the point #p# and the center of the circle.

The center of the circle #x^2 + y^2 = 20# is a #(0, 0)#, So the slope #m'# is

#m' = (y_1 - y_2)/(x_1 - x_2)#

#=> m' = (0 - (-4))/(0 - 2)#

#=> m' = 4/-2#

#=> m' = -2#

To get the slope of the tangent line (which is perpendicular), we get the negative reciprocal of the slope we computed earlier.

#m = -1/(m')#

#=> m = -1/(-2)#

#=> m = 1/2#

Now let's get the y-intercept. The equation of a line in slope-intercept form is

#y = mx + b#

To get the y-intercept, let us substitute the coordinate values of an point that we know is on the line. For the tangent line, we know it passes through #(2, -4)#

#y = mx + b#

#=> -4 = (1/2)(2) + b#
#=> -4 = 1 + b#
#=> b = -3#

Therefore, the equation of the line tangent to the point (2, -4) is

#y = 1/2x - 3#

If a line #k# is tangent to the circle at point #(k_x, k_y)# is parallel to another line #k'# that is tangent to the circle at point #(k_x', k_y')#, then either #k'# coincides with #k# (i.e. #k' = k#), or #k'# is on the exact other side of the circle. Let us leave the trivial coincidental lines case and focus on the other case.

What is the #(k_x, k_y)'s# relationship with #(k_x', k_y')#?
Since the parallel tangent lines are exactly on opposite sides of the circle, both points are endpoints of a circles diameter. We can use the midpoint formula to find #(k_x', k_y')#

#m_x = (k_x + k_x')/2#
#m_y = (k_y + k_y')/2#

Since the center of the circle is also the midpoint of the diameter, we have #m = (0, 0)#

#=> m_x = 0 = (2 + k_x')/2 = 2 + k_x'#

#=> k_x' = -2#

-

#=> m_y = 0 = (-4 + k_y')/2 = -4 + k_y'#

#=> k_y' = 4#

Hence, the other line is tangent to the circle at point #(-2, 4)#.
Since the other line is parallel to the first tangent line, the slope of the 2 lines should be the same.

#y = 1/2x + b#

Again, we need to find the #y#-intercept. Let's use the point of tangency to find it

#4 = 1/2(-2) + b#
#4 = -1 + b#
#b = 5#

Hence, the equation of the tangent line that is parallel to the tangent line at point #(2, -4)# is

#y = 1/2x + 5#