What is the equation of the tangent line of the one-family of curve x^2 + y^2 = 25 at

P(4, 3) ??

help please

Use implicit differentiation to get

2x+2yy'=0
from which
y'=-x/y=-4/3
The tangent line is therefore
(y-3)=(-4/3)(x-4)
Simplify if necessary.

I have 3 choices for that

A 4x -3y + 25 = 0
B 4x + 3y - 25 = 0
C 3x + 4y -24 = 0

i dunno what is the correct answer

Expand (y-3)=(-4/3)(x-4)

and move all the terms to the left hand side, and you should find that the equation matches one of the answers.

If it does not, post what you've got.

To find the equation of the tangent line at a given point on a curve, we can use the derivative. The derivative will give us the slope of the tangent line at any point on the curve.

First, let's differentiate the equation of the curve with respect to x:

d/dx (x^2 + y^2) = d/dx (25)

Using the chain rule, we differentiate y^2 with respect to y and multiply by dy/dx:

2x + 2y * dy/dx = 0

Now let's find dy/dx by rearranging the equation:

dy/dx = -2x / (2y)

Now we have the slope of the tangent line at any point on the curve. To find the slope at point P(4, 3), substitute the values x = 4 and y = 3 into the equation:

dy/dx = -2*4 / (2*3) = -8/6 = -4/3

So, the slope of the tangent line at P(4, 3) is -4/3. Now let's use the point-slope form of the equation of a line to find the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values x1 = 4, y1 = 3, and m = -4/3, we get:

y - 3 = (-4/3)(x - 4)

To simplify, we can multiply both sides of the equation by 3 to eliminate the fraction:

3y - 9 = -4(x - 4)

Expanding the equation:

3y - 9 = -4x + 16

Now, let's rearrange the equation to get it into the standard form, y = mx + c:

4x + 3y = 25

So, the equation of the tangent line to the one-family of curves x^2 + y^2 = 25 at point P(4, 3) is 4x + 3y = 25.