Question

Find the equation of a line tangent to the circle x2+(y−3)2=34 at the point (5, 0).

Answer 1

Answer:

Step-by-step explanation:

In order to find the equation of the line tangent to that circle, we have to find the derivative by implicit differentiation which will give us the slope formula of that tangent line.  Let's begin by expanding through the parenthesis to get the standard form of the circle:

$x^2+y^2-6y+9=34$

Moving the 9 to the other side since the derivative of a constant is 0 gives us

$x^2+y^2-6y=25$

By implicit differentation, the derivative of that function is

$2x+2y\frac{dy}{dx}-6\frac{dy}{dx}=0$

To find the derivative, we have to solve for dy/dx:

$2y\frac{dy}{dx}-6\frac{dy}{dx}=-2x$

Factor out the dy/dx to get:

$\frac{dy}{dx}(2y-6)=-2x$

Now divide to get your slope formula (first derivative):

$\frac{dy}{dx}=\frac{-2x}{2y-6}$

Now we can sub in the x and y values from the coodinate to get the slope of that tangent line:

$\frac{dy}{dx}=\frac{-2(5)}{2(0)-6}=\frac{-10}{-6}=\frac{5}{3}$

So now that have the slope, we can use the point-slope form of a line to write the equation of the tangent line.  The point-slop form of a line is:

y-y₁ = m(x-x₁)

Filling in we get:

y - 0 = 5/3(x - 5) so the equation of the tangent line is:

$y=\frac{5}{3}x-\frac{25}{3}$

Good luck in your calculus class!

Answer 2

The equation to the line tangent is:

$y = \frac{5}{3}(x - 5)$

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The equation of line, in point-slope form, is:

$y - y_0 = m(x - x_0)$

In which:

• The point is $(x_0, y_0)$.
• The slope is m.

In this problem:

• Point (5,0), thus $x_0 = 5, y_0 = 0$.
• The slope is the derivative at the point, thus, we apply implicit differentiation.

$2x\frac{dx}{dx} + 2(y-3)\frac{dy}{dx} = 0$

$2(5) + 2(0 - 3)\frac{dy}{dx} = 0$

$6\frac{dy}{dx} = 10$

$m = \frac{dy}{dx} = \frac{10}{6} = \frac{5}{3}$

Thus, the equation is:

$y = \frac{5}{3}(x - 5)$

A similar problem is given at https://brainly.com/question/22426360