The center of a circle is the point D(-2,1), and Q(1,-3) is a point on the circle. Find the equation of the line tangent to the circle at the point Q
EDIT:
The slope of the radius is -4/3 if I didn't make an error in my calculation. Also I am in algebra 2 and this is supposed to be review… I am stumped, I have never seen this before.
Best Answer
Consider the point-slope formula:
$$\left( y-y_1 \right)=m\left( x-x_1 \right).$$
This is nothing more than a consequence of the definition of slope. More specifically,
$$m=\frac{\left( y_2-y_1 \right)}{\left( x_2-x_1 \right)}\Rightarrow \left( y_2-y_1 \right)=m\left( x_2-x_1 \right).$$
If we are in a situation where we already know the slope $m$ and we already know a point $( x,y )$, we can drop one of the points in that last form (the $_2$ in $x_2$ and $y_2$ for example) and write the point-slope formula as above.
Using the point-slope form $$(y−y_1)=m(x−x_1),$$ with $m=\frac{3}{4}$, and the point $(1,-3)$ we substitute our given $m$, $x$, and $y$ to get. $$(y−(−3))=\frac{3}{4}(x−1).$$
If you want to convert this to the slope intercept form then the process is
\begin{align*} &(y−(−3))=\frac{3}{4}(x−1) \\ \Rightarrow & y+3=\frac{3}{4}x-\frac{3}{4} \\ \Rightarrow & y=\frac{3}{4}x-\frac{3}{4}-3 \\ \Rightarrow & y=\frac{3}{4}x-\frac{15}{4}. \end{align*}
Note that I still recommend my original solution where we start with the slope-intercept form. It is much faster (and seems much more sensible to me), although I have noticed a strange reluctance to demonstrate that method directly in the textbook (Bittinger) that I am currently using for my classes.