tag:blogger.com,1999:blog-49623322828247694332011-10-04T13:09:33.976-07:00Ketut Widartanoreply@blogger.comBlogger11100tag:blogger.com,1999:blog-4962332282824769433.post-29903743906512905712011-09-21T13:42:00.000-07:002011-09-21T13:42:41.490-07:00Sample Question :<br /> Consider the circle of radius 5 centered at (0,0). Find an equation of the line tangent to the circle at the point (3,4).<br /><br /> <br /> Answer :<br /><br />The line tangent to a circle is also perpendicular to the radius drawn to the point of tangency. So finding the slope of the radius out to the tangent line is easy in this case, since you're starting at the origin and going out to (3.4): rise/run, or 4/3. <br /><br />The tangent line will be perpendicular to the radius, so it's slope will be a negative reciprocal: -3/4<br /><br />Now use y = mx + b, substitute in -3/4 for m, and (3,4) for x and y to find b:<br /><br />4 = -3/4(3) + b<br />4 = -9/4 + b<br />b = 25/4<br /><br />So the equ'n of the tangent line is y = -(3/4)x + 25/4.<br /><br /><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/4962332282824769433-2990374390651290571?l=textualtangents.blogspot.com' alt='' /></div>Ketut Widartanoreply@blogger.com0