# Find the equation of the line tangent to the curve x^2+y^2=25 ; (-3,4) at given point.

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### 1 Answer

You should remember the equation of the tangent line to a curve whose equation is `y = f(x), ` at a point `(x_0,y_0), ` such that:

`y - f(x_0) = f'(x_0)(x - x_0)`

You need to write the equation of the curve in terms of x such that:

`y^2 = 25 - x^2 => y = +-sqrt(25-x^2)`

You need to differentiate the function with respect to x, such that:

`f'(x) = ((25-x^2)')/(2sqrt(25-x^2))`

`f'(x) = -2x/(2sqrt(25-x^2)) => f'(x) = -x/(sqrt(25-x^2)) `

You need to evaluate `f'(-3)` such that:

`f'(-3) = 3/(sqrt(25-9)) => f'(-3) = 3/sqrt16 => f'(-3) = 3/4`

Since the problem provides `f(-3) = 4` , you may write the equation of the tangent line such that:

`y - 4 = (3/4)(x + 3) => y = 3x/4 + 9/4 + 4 => y = 3x/4 + 25/4`

**Hence, evaluating the equation of the tangent line, at the point `(-3,4), ` to the given curve, yields `y = 3x/4 + 25/4.` **