What Are the Equations of Tangent Lines to a Circle Without Using Calculus?

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Homework Help Overview

The problem involves finding the equations of tangent lines to a circle defined by the equation x² + (y + 1)² = 1, which intersect at the point (0, 1). The challenge is to solve this without using calculus or derivatives.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial setup of the problem, including the form of the tangent lines and their intersection with the circle. There is an exploration of the implications of the quadratic formula in relation to the number of solutions, particularly in the context of tangents. Some participants also question the validity of the point (0, 1) being outside the circle.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have offered insights into the symmetry of the tangent lines and their equations, while others have raised concerns about the initial conditions and assumptions regarding the circle's geometry.

Contextual Notes

There is a noted constraint that the problem must be solved without calculus, which has led to discussions about the implications of this restriction on the methods used. Additionally, the point (0, 1) is identified as not lying on the circle, which may affect the approach to finding the tangent lines.

hawk320
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Homework Statement


You have a circle with the equation x^{2} + (y + 1)^{2} = 1. You can draw to two tangent lines to that circle that intersect the point (0,1) What are the equations of these lines? And you can't use any calculus, derivatives and the like.


Homework Equations


y=mx+b
quadratic formula
x^{2} + (y + 1)^{2} = 1

The Attempt at a Solution


Well you can begin by knowing that the lines y-int will be 1 so y=mx+1. Then you can solve the equation for the circle for y which gives you y = -1 \pm \sqrt{1-x^{2}}. Then you can set that equation equal to 0 and get (after factoring) (1+m ^{2}) * x^{2} + 4mx + 3 = 0. Then you can plug this into the quadratic formula to get your x, but there I get stuck. i try to plug that back into y = mx +1 but I don't know what I am looking for.
 
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Tangent only touches the circle right? So if you solve, you should get only one solution.

For ax^2+bx+c=0 if b^2-4ac=0 how many solutions are there?
 
the data is correct, the point (0,1) does not lie on the circle the highest point of the circle is (0,0)
 
Well you have equation of the tangent lines: y=mx+1. Note that by symmetry, you can assume that the x-coordinate of the point where the lines touch the circle are x and -x and their y coordinates are the same So, you have y=mx+1 for both equations:
(1)y=m1x + 1
(2)y=m2(-x) + 1

You also have the equation of the circle. You should see how to continue from here.

EDIT: This thread shouldn't be in this forum since you can't use calculus to solve the problem.
 

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