From the point (0,-5), tangent lines are drawn to the circle
(x-3)^2 + y^2 = 4. Find the slope of each tangent. I want the steps only.
Thank you.
Firstly, this question is from a chapter in my precalculus text titled Analytic Geometry. Secondly, this is not homework for me. I am 48 years old. I graduated college 19 years ago.
Thirdly, I am reviewing precalculus on my own not preparing for any particular test. I love math and always enjoy a good review or learning new material. There is nothing else to do it than that.
I can take the derivative of both sides but this is a precalculus question. I am to find the slope using the following formula:
r = |mx - b - y|/sqrt{1 + m^2}, Where r is the radius of the circle, m is the slope, x and y are the coordinates of a point in the form (x,y) and b is the y-intercept in the formula y = mx + b. The point-slope and midpoint formulas also play a role in solving this question. Can you help me find the slope in this case without applying calculus?
Let the circle be with center O, its equation is ( x-3)^2 + y^2 = 4. Thus we have the coordinates of the center of circle O ( 3,0)
Let P ( h,k) be a point on the circle where the tangent to the circle from A ( 0, -5 ) meets the circle.
Equation of a line through A is y = mx - 5 ----- [1]
Slope of OP = k/(h-3) ----- [2]
But AP is perpendicular to OP thus their product should be -1. Hence we get
mk/(h-3) = -1 OR mk = 3 - h ------ [3]
Lastly P lies on the circle thus we have
( h-3)^2 + k^2 = 4. ------- [4]
I am sure now you can help yourself.