Question 1171761
<font face="Times New Roman" size="+2">


First, it would be helpful to know the question that you are trying to answer.  I suspect that you want to know the radius of the circle perhaps so that you can derive the circle's equation.  However, "the line above" is a bit non-specific, so the best I can do is describe the procedure.


The underlying facts that allow you to solve this are that a line through the center of a circle and through a point of tangency is perpendicular to the tangent line and that perpendicular lines have reciprocal slopes.


I don't know what you know about the "line above."  If you have an equation of the line, put it in Slope-intercept form to find the slope by inspection; if you have two points on the line, use the Slope formula, or if you are just given the slope, use it.


Find the negative reciprocal of the slope of the given line.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_1\ =\ -\frac{1}{m_2}]


Use the Point-Slope form to find an equation of the line perpendicular to the given line through the circle's given center.


Point-Slope Form


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1)]



The two equations, the given line and the perpendicular through the center, form a 2X2 system of equations.  Solve the system for the point of intersection, which is the point of tangency.


Use the distance formula to find the distance from the center of the circle to the point of tangency.


Distance Formula


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \sqrt{ (x_1\,-\,x_2)^2\ +\ (y_1\,-\,y_2)^2}]


Hint: If your goal is to write the circle equation, you don't actually have to take the square root in the distance formula because the circle equation requires *[tex \Large r^2].

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
I > Ø
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>