Find an equation of the tangent line to the circle x^2+y^2=25 at the point (4,-3)
at (x,y) the slope is -y/x
So, at (4,-3) the slope is 3/4.
To find the equation of the tangent line to a circle at a given point, we need to find the gradient (slope) of the tangent line and the coordinates of the point of tangency.
Step 1: Find the gradient of the tangent line
The gradient of a circle at any given point is given by the negative reciprocal of the gradient of the radius passing through that point.
The equation of the given circle is x^2 + y^2 = 25, which can be rewritten as y = sqrt(25-x^2) or y = -sqrt(25-x^2) (think about the positive and negative square root).
In this case, we have (4,-3) as the point of tangency. Let's find the slope of the radius at this point.
Differentiating the equation y = sqrt(25-x^2), we get:
dy/dx = -(1/2)*(25-x^2)^(-1/2) * (-2x)
= x/(sqrt(25-x^2))
Evaluating this derivative at x=4 (since we want the tangent line at the point (4,-3)):
dy/dx = 4/(sqrt(25-16))
= 4/3
So the slope of the radius at the point (4,-3) is 4/3.
Step 2: Find the point of tangency
We already know that the point of tangency is (4,-3), so no additional calculations are needed for this step.
Step 3: Use the point-slope form to find the equation of the tangent line
We have the point of tangency (4,-3) and the slope of the tangent line (4/3). Now we can use these values in the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1),
where (x1, y1) is the point of tangency and m is the slope of the tangent line.
Substituting the values, we have:
y - (-3) = (4/3)(x - 4)
y + 3 = (4/3)x - 16/3
y = (4/3)x - 16/3 - 3
y = (4/3)x - 16/3 - 9/3
y = (4/3)x - 25/3
Therefore, the equation of the tangent line to the circle x^2 + y^2 = 25 at the point (4, -3) is y = (4/3)x - 25/3.
Oops. The slope is -x/y.
SO, it's 4/3