Find an equation of the line that is tangent to the circle x^2 + y^2 = 25 at the point P(−3, −4).
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.
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
To find the equation of a line tangent to a circle at a given point, we need to find the slope of the tangent line.
Step 1: Find the derivative of the equation of the circle.
The equation of the circle is x^2 + y^2 = 25.
Differentiating both sides with respect to x gives us:
2x + 2y * dy/dx = 0
Simplifying this equation, we get:
dy/dx = -x/y
Step 2: Evaluate the derivative at the given point P(-3, -4).
Substituting the coordinates of P into the derivative equation, we get:
dy/dx = -(-3)/(-4) = 3/4
Step 3: Use the point-slope form of a line to find the equation of the tangent line.
The point-slope form of a line is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
Using the point P(-3, -4) and the slope 3/4, we can write the equation of the tangent line as:
y - (-4) = (3/4)(x - (-3))
Simplifying the equation, we get:
y + 4 = (3/4)(x + 3)
This is the equation of the tangent line to the circle x^2 + y^2 = 25 at the point P(-3, -4).