<p><a href="http://i.im%5B/url%5D">http://i.im</a> gur.com/XjetCdp.png (no spaces)</p>
<p>I actually got this problem right, by plugging -4 for x and getting y as 3, then eyeballing (0,7) to be another point on the line, therefore getting the slope as 4/3, but what's the way to do it algebraically?</p>
<p>Well you know that the radius of the circle is 5 since it’s the square root of 25. And you know (or should know) that a line tangent to a circle is perpendicular to the radius of the circle. That’s the key: You have to find the slope of the radius touching line l to be able to find the slope of line l since perpendicular lines have negative reciprocal slopes.
Point A is the leading gate to that answer. The x-coordinate of point A is -4. You want to find the y-coordinate. Since every radius in that circle is 5 and the center of the circle is 0,0 you can use the distance formula to find the y-coordinate due to that radius being a line. 5 is the distance so you plug that in to equal sqrt((-4-0)^2 + (y-0)^2). You find y to be 3. Now you use the slope formula to find the slope of that radius which gives you 3/-4. So, the slope of line l is 4/3.</p>
<p>[P.S. It might seem like a lot but that’s only because writing the process down ins lenghty]</p>
<p>The “mathematically correct” way to do it would be to find the slope of the radius of the circle - use the points (0,0) and (-4,3) to get m = -3/4. </p>
<p>A tangent line to a circle is perpendicular to the corresponding radius. Therefore the slope of line l is the negative reciprocal of the slope of the radius: 4/3.</p>
<p>Thanks
10char</p>