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 September 9th, 2008, 10:17 AM #1 Newbie   Joined: Sep 2008 Posts: 1 Thanks: 0 A problem concerning a tangent line and a circle First of all hi to everybody, I'm new to these forums, but i was previously quite active on another math-related forum (which is sadly almost dead nowadays). So here's the problem I currently can't solve, For what values of k is the straight line y = kx + 1 a tangent to the circle with the origin at (5, 1) and a radius of 3.
 September 9th, 2008, 02:45 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: A problem concerning a tangent line and a circle A circle of radius r, centre $\displaystyle (x_0,y_0)$ can be described by the equation $\displaystyle (x-x_0)^2+(y-y_0)^2=r^2.$ When you have simplified this for the circle given in the problem, you need to find values of x, y and k such that (i) The line and the circle intersect (i.e. x, y and k satisfy both the equation for the line and the equation for the circle); (ii) The slope of the line and the circle are equal at this point (i.e. $\displaystyle \frac{dy}{dx}$ is equal on the line and the circle when evaluated for these values of x, y and k.) Last edited by skipjack; February 28th, 2018 at 12:57 PM.
 September 9th, 2008, 04:56 PM #3 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: A problem concerning a tangent line and a circle If not differentiating yet, the second relation can be found from the fact that the line from the point of tangency to the center is perpendicular to the tangent line y = kx + 1.
 September 10th, 2008, 01:02 PM #4 Global Moderator   Joined: May 2007 Posts: 6,756 Thanks: 696 Re: A problem concerning a tangent line and a circle Compute the intersection of the line (as a function of k) with the circle. The solution set will consist of one of the following: two different intersections, two identical intersections, or no intersection. The k's (there should be 2) which lead to two identical intersections are the tangent lines.
 September 10th, 2008, 03:07 PM #5 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 Re: A problem concerning a tangent line and a circle For a circle of radius $\displaystyle R$ centered at $\displaystyle (x_0, \, y_0)$ to have a tangent of the form $\displaystyle y = k x + l$ you need to have the following condition fulfilled: $\displaystyle (1+k^2)R^2=(l+k x_0-y_0)^2$ Solving for $\displaystyle k$ yields: $\displaystyle k_{1,2} = \frac{x_0 (l - y_0)\pm R\sqrt{x_0^2+y_0^2-R^2+l^2-2 l y_0}}{R^2-x_0^2}$ In your problem, you have $\displaystyle l = 2$, $\displaystyle x_0= 2$, $\displaystyle y _0= 1$ and $\displaystyle R = 2$, so the possible values for $\displaystyle k$ are: $\displaystyle k = \pm \frac{3}{4}$ If you are interested, the tangents touch the circle at points $\displaystyle \left(\frac{16}{5},\, \frac{17}{5}\right)$ and $\displaystyle \left(\frac{16}{5},\, -\frac{7}{5}\right)$. Last edited by skipjack; February 28th, 2018 at 12:55 PM.
 September 10th, 2008, 03:28 PM #6 Senior Member   Joined: Jul 2008 Posts: 895 Thanks: 0 Re: A problem concerning a tangent line and a circle The solution may be simpler because of the geometry. The tangent lines intersect at the point (0,1), and are perpendicular to the radius of the given circle. The point(s) of tangency (x,y) and ((0,1) and (5,1) form a right triangle. So, (0,1) and (5,1) form the diameter of an intersecting circle of radius 2.5. center (2.5,1).

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# find the value of k such that the straight line may touch the circle

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