so that so that . That is the equation of a circle with center (2, 0) and radius 1. The problem ask for you to find two equations through the origin (so y= mx for some m) that are tangent to the circle. It should be clear, geometrically, that one is above the x-axis and the other below.

One way to do this is to argue that the line must be tangent to the circle and so replacing y by mx, [must](x- 2)^2+ (mx)^2= 1[/tex], must have adoubleroot. That is a quadratic equation so must have zero, one, or two solutions. Since a tangent touches the circle, it cannot have 0 solutions. Since a tangent does not passthroughthe circle, it does not cross it twice and so that quadratic equation must have only one solution. You can set thediscriminantof the equation equal to 0 to determine m.

Another way to do this is to use the fact that a tangent to a circle is perpendicular to a radius. So the origin, (0, 0), the center of the circle, (2, 0), and the point, (x, y), where the line is tangent to the circle form a right triangle with the x-axis, from (0, 0) to (3, 0) as hypontenuse. That is, since the distance from (0, 0) to (2, 0) is 2 and the circle has radius 1. That, together with the equation of the circle gives you two equations to solve for x and y. Once you know the point, (x, y), where the line is tangent to the circle, as well as (0, 0) you can write down the equation of the line.