Question

How do you find the equation of the normal to the circle \[{x^2} + {y^2} = 40\] at the point \[\left( {6,2} \right)\] ?

Answer 1

Hint: Here in this question, we have to find the equation of the normal to the circle. 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:
By circle properties, since the tangent is perpendicular to the radius of the circle at the point \[\left( {6,2} \right)\], the normal, which is perpendicular to the tangent, must be parallel to the radius. Hence the two points are \[\left( {6,2} \right)\] and \[\left( {0,0} \right)\]. Consider the equation of circle
\[{x^2} + {y^2} = 40\]-------(1)
Now, we have to find the equation of the normal to the circle \[{x^2} + {y^2} = 40\] at the point \[\left( {6,2} \right)\] 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 \[{x_1} = 6\], \[{x_2} = 0\], \[{y_1} = 2\] and \[{y_2} = 0\] on substituting this in formula, then
 \[m = \dfrac{{0 - 2}}{{0 - 6}}\]
\[ \Rightarrow \,\,\,m = \dfrac{{ - 2}}{{ - 6}}\]
On simplification, we get
\[m = \dfrac{1}{3}\]
Now we get the gradient or slope of the line which passes through the points \[\left( {6,2} \right)\]. Substitute the slope m and the point \[\left( {{x_1},{y_1}} \right) = \left( {6,2} \right)\] in the point slope formula.Consider the equation (2)
\[y - {y_1} = m\left( {x - {x_1}} \right)\]
Where \[m = \dfrac{1}{3}\], \[{x_1} = 6\] and \[{y_1} = 2\] on substitution, we get
\[y - 2 = \dfrac{1}{3}\left( {x - 6} \right)\]
\[ \Rightarrow \,\,y - 2 = \dfrac{1}{3}x - \dfrac{1}{3}\left( 6 \right)\]
\[ \Rightarrow \,\,y - 2 = \dfrac{1}{3}x - 2\]
Add 2 on both side, then
\[y - 2 + 2 = \dfrac{1}{3}x - 2 + 2\]
On simplification, we get
\[\therefore\,\,y = \dfrac{1}{3}x\]

Hence, the equation of the normal to the circle \[{x^2} + {y^2} = 40\] at the point \[\left( {6,2} \right)\] is \[y = \dfrac{1}{3}x\].

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.