Question

How do you find the equation for the normal line to \[{x^2} + {y^2} = 9\] through \[\left( {0,3} \right)\] ?

Answer 1

Hint: Here in this question, we have to find the equation of the normal line. Find the equation by using the Point-Slope formula \[y - {y_1} = m\left( {x - {x_1}} \right)\] before finding the equation first we have to find the slope using the formula \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\] . On simplification to the point-slope formula we get the required solution.

Complete step by step solution:
The given equation represents \[{x^2} + {y^2} = 9\] the equation of Circle \[{x^2} + {y^2} = {r^2}\] . Then by circle properties, since the tangent is perpendicular to the radius of the circle at the point \[\left( {0,3} \right)\] , the normal, which is perpendicular to the tangent, must be parallel to the radius.
Hence the two points is \[\left( {0,3} \right)\] and \[\left( {0,0} \right)\] .
Consider the equation of circle
 \[{x^2} + {y^2} = 9\] -------(1)
Now, we have to find the equation of the normal to the circle \[{x^2} + {y^2} = 9\] at the point \[\left( {0,3} \right)\] by using the slope-point formula \[y - {y_1} = m\left( {x - {x_1}} \right)\] -------(2)
Before this, find the slope \[m\] in point-slope formula by using the formula \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\]
Where \[{x_1} = 0\] , \[{x_2} = 0\] , \[{y_1} = 3\] and \[{y_2} = 0\] on substituting this in formula, then
  \[ \Rightarrow m = \dfrac{{0 - 3}}{{0 - 0}}\]
 \[ \Rightarrow m = \dfrac{{ - 3}}{0}\]
On simplification, we get
 \[ \Rightarrow m = \infty \]
The slope m is undefined.
Now we get the gradient or slope of the line which passes through the points \[\left( {0,3} \right)\] .
Substitute the slope m and the point \[\left( {{x_1},{y_1}} \right) = \left( {0,3} \right)\] in the point slope formula.
Consider the equation (2)
 \[y - {y_1} = m\left( {x - {x_1}} \right)\]
Where \[m = \infty \] , \[{x_1} = 0\] and \[{y_1} = 3\] on substitution, we get
 \[ \Rightarrow y - 3 = \infty \left( {x - 0} \right)\]
Hence, the normal line will be vertical at \[\left( {0,3} \right)\] and will be of the form \[x = a\] .
Since the point of intersection is \[x = 0\] , the normal line will have equation \[x = 0\] . So the \[y\] axis is the normal line at that point.

So, the correct answer is “ \[x = 0\] ”.

Note: The concept of the equation of normal comes under the concept of application of derivatives. Here the differentiation is applied on the equation of line or curve and then substitute the value of x. We should know about the general equation of a line. Hence these types of problems are solved by the above procedure.