How do you find the equations for the normal line to #x^2+y^2=9# through (0,3)?

Question

Paisley Johnson · Accepted Answer

#x= 0#

We find the derivative.

#2x + 2y(dy/dx) = 0#

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

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

#dy/dx = -x/y#

We now determine the slope of the tangent.

#m_"tangent" = -0/3#

#m_"tangent" = 0#

The normal line is perpendicular to the tangent, so the slope of the normal line will be #-1/k# of that of the tangent, where #k# is the slope of the tangent.

#m_"normal" = O/#

Hence, the normal line will be vertical at #(0, 3)# and will be of the form #x = a#.

Since the point of intersection is #x = 0#, the normal line will have equation #x= 0#.

Hopefully this helps!

Samantha Adkison · Answer

Here is a solution using geometry.

#x^2+y^2=9# is a circle centered at #(0,0)# with radius #3#. The point #(0,3)# is on the #y#-axis. The line segment joining #(0,0)# and #(0,3)# is a radius of the circle and the radius is perpendicular to the tangent. So the #y# axis is the normal line at that point.

Isaiah Allen · Answer

Here is a solution using calculus, but without implicit differentiation.

#x^2+y^2=9# #y = +-sqrt(9-x^2)# The point #(0,3)# is a solution to #y = sqrt(9-x^2)# so we are interested in the function #f(x) = sqrt(9-x^2)#. Differentiate using the chain rule to get #f'(x) = 1/(2sqrt(9-x^2)) * (-2x) = (-x)/sqrt(9-x^2)# At #(0,3)#, the tangent line has slope #f'(0) = 0#. So the tangent line is a horizontal line and the normal line is vertical. The vertical line through #(0,3)# has equation #x=0#.

Luke Alvarez · Answer

undefined To find the equation of the normal line to the circle $x^2 + y^2 = 9$ through the point $(0,3)$, follow these steps:

1. First, find the derivative of the circle's equation.
2. Then, find the slope of the tangent line at the point of tangency (which is the point of intersection between the circle and the normal line).
3. Use the negative reciprocal of this slope to find the slope of the normal line.
4. Finally, use the point-slope form to write the equation of the normal line.

Here are the steps in more detail:

1. The equation of the circle is $x^2 + y^2 = 9$. Taking the derivative with respect to $x$, we get:
$$2x + 2y\frac{{dy}}{{dx}} = 0$$
$$y\frac{{dy}}{{dx}} = -x$$
$$ \frac{{dy}}{{dx}} = -\frac{{x}}{{y}}$$

2. At the point $(0,3)$, substitute $x = 0$ and $y = 3$ into the derivative to find the slope of the tangent line:
$$m_{\text{tangent}} = -\frac{{0}}{{3}} = 0$$

3. The slope of the normal line is the negative reciprocal of the slope of the tangent line:
$$m_{\text{normal}} = -\frac{1}{{m_{\text{tangent}}}} = \text{undefined}$$

4. Since the slope of the normal line is undefined, the normal line is vertical and passes through the point $(0,3)$. Therefore, its equation is $x = 0$.