How do you write an equation for a circle tangent to the line 5x + y = 3 at the point (2,-7) and the center is on the line x – 2y= 19?

1 Answer
Oct 25, 2016

Answer:

Please see the explanation. The equation is:
#(x - 7)^2 + (y - -6)^2 = (sqrt(26))^2#

Explanation:

Write the equation of the tangent line in slope-intercept form so that we may observe its slope.

#y = -5x + 3#

The slope #m = -5#

Because radius of the circle at point (2, -7) will be perpendicular to the tangent line, the slope of the radius is the negative reciprocal of the slope of the tangent line, #1/5#, Write the equation of the line through the point with this slope:

#-7 = 1/5(2) + b#

#b = -7 -2/5#

#b = -37/5#

#y = 1/5x - 37/5# [1]

The center #(h, k)# will be at the intersection of equation [1] and the line:

#x - 2y = 19# [2]

Substitute equation [1] into equation [2]:

#x - 2(1/5x - 37/5) = 19#

#x - 2/5x + 74/5 = 19#

#3/5x = 21/5#

#x = 7#

Substituting for x in equation [2]:

#7 - 2y = 19#

#- 2y = 12#

#y = -6#

Substitute the center #(7, -6)# and the point #(2, -7)# into the equation of a circle:

#(2 - 7)^2 + (-7 - -6)^2 = r^2#

#r = sqrt(26)