Step 1: Recognize that a tangent to a circle makes a right angle with the radius at the point of tangency (hereinafter referred to as point T). Hence, the line segment between the center of the circle and the point (4,6) (hereinafter referred to as point A) is the hypotenuse of a right triangle where the legs are the radius at point T and the segement between point T and point A. Also note that your circle is centered at the origin, point O.
Step 2: Determine the length of segment TA. Use the distance formula to calculate the measure of OA. Determine the radius of the given circle by inspection of the equation. Use these values for the hypotenuse and one leg of the right triangle described in Step 1 to calculate the measure of TA.
Step 3: Notice that the segment TA is the radius of a circle centered at A that intersects the given circle at T. It is here that we realize that this problem will have two solutions because this new circle does, in fact, intersect the original circle at two points.
Step 4: Having the radius and center of this new circle, write the equation of this new circle.
Step 5: Solve the 2X2 quadratic system consisting of the equations of the two circles. It will be convenient to write both equations as functions of
Step 6: Using the two-point form of an equation of a line, derive the two desired equations -- one containing the segment
As you can see from the description of the solution, this is a complex problem with a labor intensive and difficult solution, and not something I care to do for free. Please write back for a quotation for a complete solution including a full explanation and graphical illustration, if desired.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
