Question 194785
First, answer the question: "Is the point (-7,1)
on the circle?"
{{{x^2 + y^2 = 50}}}
{{{(-7)^2 + 1^2 = 50}}}
{{{49 + 1 = 50}}}
{{{50 = 50}}} Yes
The center of the circle is the origin.
Next find the equation of the line  that
contains the points (0,0) and (-7,1)
{{{(y - 1) / (x - (-7)) = (0 - 1) / (0 - (-7))}}}
{{{(y - 1) / (x + 7) = -1 / 7}}}
Multiply both sides by {{{7*(x + 7)}}}
{{{7*(y - 1) = -(x + 7)}}}
{{{7y - 7 = -x - 7}}}
{{{7y = -x}}}
{{{y = -(1/7)*x}}}
This is of the form {{{y = mx}}}, and {{{m}}} is the slope
Any line perpendicular to this one will have slope
{{{-(1/m)}}}, in this case, {{{-(1/(-(1/7))) = 7}}}
So, the line tangent to the circle at (-7,1) will have 
slope = {{{7}}}.
{{{(y - 1) / (x - (-7)) = 7}}}
{{{y - 1 = 7*(x + 7)}}}
{{{y - 1 = 7x + 49}}}
{{{y = 7x + 50}}} answer
I'll plot the circle and the line
{{{ graph( 500, 500, -60, 60, -60, 60, sqrt(50 - x^2), -sqrt(50 - x^2), 7x + 50)}}}