The line y=mx is a tangent to the circle x^2+(y-2)^2=2.

Question

Find algebraically the values of m and the points of contact of the tangents to the circle.

Answer 1

 

x²+(y-2)²=2   is a circle with its center C at (0,2) and r=√(2)    ( from  (x-h)²+(y-k)²=r² , std form of a circle )

 

y=mx meets this circle tangentially at point A in Q1,  and point B in Q2.   A & B are symmetrical about the y-axis because the circle is 

centered on the y-axis, so lets find A, then B will be obvious.   Also note y=mx passes through the origin O(0,0).

 

If you draw a triangle OAC you will find |OC|=2, |CA|=√(2)  and, noting the tangent meets the circle at 90^o, you can

solve for |OA|=√(2).   Now a triangle drawn from A down to the x-axis and over to the origin is also a 45-45-90 triangle

and A(a_x,a_y)  can be easily found (cos(45^o) = √(2)/2,  sin(45^o) = √(2)/2)...

 

A is therefore at (1,1) and the value of m=1.  

B is at (-1,1) and m=-1 for the other line.

 

 

There are probably more efficient methods to find m, A, and B.  I just followed the first process that came to mind.