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.

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- October 28th 2013, 07:10 PMnycmathFind Slope of Each Tangent
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. - October 29th 2013, 12:13 AMchiroRe: Find Slope of Each Tangent
Hey nycmath.

We can't you the complete answer (at least that's the idea and intent of homework help), but you should start off by differentiating both sides and note that d/dx(y^2)= 2y*dy/dx through the chain rule. - October 29th 2013, 06:16 AMnycmathRe: Find Slope of Each Tangent

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? - October 29th 2013, 05:07 PMchiroRe: Find Slope of Each Tangent
Hint: The slope for a circle is tan(theta) = m = -y/x given a point (x,y) on the 2D plane.

- October 31st 2013, 10:22 PMnycmathRe: Find Slope of Each Tangent
I will have to work on this one some more.

- October 31st 2013, 10:45 PMibduttRe: Find Slope of Each Tangent
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.