Asked by Ashton
                Find an equation of the line that is tangent to the circle x^2 + y^2 = 25 at the point P(−3, −4).
            
            
        Answers
                    Answered by
            bobpursley
            
    Algebra?
I cant think of a way to determine the tangent to the circle except with calculus
2x dx=-2y dy
dy/dx=-x/y=m=-3/4
y=-3/4 x + b
find b
-4=-3/4 (-3) + b
-4+9/4 = b
so you now have it.
    
I cant think of a way to determine the tangent to the circle except with calculus
2x dx=-2y dy
dy/dx=-x/y=m=-3/4
y=-3/4 x + b
find b
-4=-3/4 (-3) + b
-4+9/4 = b
so you now have it.
                    Answered by
            Damon
            
    First draw the picture, center at (0,0) and radius = 5
You see that point in Quadrant 3?
It makes a triangle with the -y axis
hypotenuse 5, bottom 3, 4 down y axis
Now draw the tangent to the circle at that point. It will be perpendicular to the radius at that point.
Since the slope of the radius there is -4/-3 = 4/3
the slope of the perpendicular is -3/4
so
y = (-3/4) x + b
put in our point to get b
-4 = (-3/4)(-3) + b
so
b = -4 - 9/4 = - 25/4
so
y = (-3/4) x - 25/4
or
4 y = -3 x -25
    
You see that point in Quadrant 3?
It makes a triangle with the -y axis
hypotenuse 5, bottom 3, 4 down y axis
Now draw the tangent to the circle at that point. It will be perpendicular to the radius at that point.
Since the slope of the radius there is -4/-3 = 4/3
the slope of the perpendicular is -3/4
so
y = (-3/4) x + b
put in our point to get b
-4 = (-3/4)(-3) + b
so
b = -4 - 9/4 = - 25/4
so
y = (-3/4) x - 25/4
or
4 y = -3 x -25
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