How to find the equation for a tangent to a circle with only one point?

How do you find the equation for a tangent to the circle #x^2+y^2-2x+10y-19=0# that passes through the point #(7,-2)#?

1 Answer
May 7, 2017

Please see the explanation.

Explanation:

Given #x^2+y^2-2x+10y-19=0# and the point #(7,-2)#

Verify that the point is on the circle:

#7^2+(-2)^2-2(7)+10(-2)-19=0#

#0=0# verified

Differentiate each term of the equation:

#(d(x^2))/dx+(d(y^2))/dx-(d(2x))/dx+(d(10y))/dx-(d(19))/dx=0#

#2x+2ydy/dx-2+10dy/dx-0=0#

Move the terms that do not contain #dy/dx# to the right side:

#2ydy/dx+10dy/dx=2-2x#

Factor the left side:

#(2y+10)dy/dx=2-2x#

Divide both sides by #2y+10#:

#dy/dx=(2-2x)/(2y+10)#

The slope, m, of the tangent line at the point #(7,-2)# is the above derivative evaluated at the point:

#m = (2-2(7))/(2(-2)+10)#

#m = (-12)/6#

#m = -2#

Use the point slope form of the equation of line:

#y = m(x - x_1)+y_1#

#y = -2(x - 7) - 2 larr# point-slope form

#y = -2x + 14 - 2#

#y = -2x + 12 larr# slope-intercept form

#2x + y = 12 larr# standard form