Question

question 2 create a line tangent to circle a passing through point b. place a point c on the tangent line at any location you choose. using the coordinates of the points, find the slope - intercept form of the equation for the tangent line bc.

Explanation:

Step1: Connect center of circle A to point B

Let the center of circle A be \(O(x_0,y_0)\). The slope of line \(OB\) is \(m_{OB}=\frac{y_B - y_0}{x_B - x_0}\) (assuming coordinates of \(O\) are \((x_0,y_0)\) and of \(B\) are \((x_B,y_B)\)).

Step2: Find slope of tangent line

Since the tangent line is perpendicular to the radius at the point of tangency, if the slope of the radius \(m_{OB}\) is known, the slope of the tangent line \(m\) (line \(BC\)) is \(m =-\frac{1}{m_{OB}}\) (product of slopes of perpendicular lines is - 1).

Step3: Use point - slope form to get slope - intercept form

The point - slope form of a line is \(y - y_B=m(x - x_B)\). Expand it: \(y - y_B=mx - mx_B\), then \(y=mx+(y_B - mx_B)\) which is in slope - intercept form \(y = mx + b\) where \(m\) is the slope and \(b=y_B - mx_B\).

Answer:

The slope - intercept form of the equation of line \(BC\) is \(y=mx+(y_B - mx_B)\) where \(m =-\frac{1}{m_{OB}}\) and \(m_{OB}=\frac{y_B - y_0}{x_B - x_0}\) (coordinates of center of circle \(O\) are \((x_0,y_0)\) and of point \(B\) are \((x_B,y_B)\)).