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

How do you find the equation for the normal line to x2+y2=9{x^2} + {y^2} = 9 through (0,3)\left( {0,3} \right) ?

Explanation

Solution

Hint : Here in this question, we have to find the equation of the normal line. Find the equation by using the Point-Slope formula yy1=m(xx1)y - {y_1} = m\left( {x - {x_1}} \right) before finding the equation first we have to find the slope using the formula m=y2y1x2x1m = \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 x2+y2=9{x^2} + {y^2} = 9 the equation of Circle x2+y2=r2{x^2} + {y^2} = {r^2} . Then by circle properties, since the tangent is perpendicular to the radius of the circle at the point (0,3)\left( {0,3} \right) , the normal, which is perpendicular to the tangent, must be parallel to the radius.
Hence the two points is (0,3)\left( {0,3} \right) and (0,0)\left( {0,0} \right) .
Consider the equation of circle
x2+y2=9{x^2} + {y^2} = 9 -------(1)
Now, we have to find the equation of the normal to the circle x2+y2=9{x^2} + {y^2} = 9 at the point (0,3)\left( {0,3} \right) by using the slope-point formula yy1=m(xx1)y - {y_1} = m\left( {x - {x_1}} \right) -------(2)
Before this, find the slope mm in point-slope formula by using the formula m=y2y1x2x1m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
Where x1=0{x_1} = 0 , x2=0{x_2} = 0 , y1=3{y_1} = 3 and y2=0{y_2} = 0 on substituting this in formula, then
m=0300\Rightarrow m = \dfrac{{0 - 3}}{{0 - 0}}
m=30\Rightarrow m = \dfrac{{ - 3}}{0}
On simplification, we get
m=\Rightarrow m = \infty
The slope m is undefined.
Now we get the gradient or slope of the line which passes through the points (0,3)\left( {0,3} \right) .
Substitute the slope m and the point (x1,y1)=(0,3)\left( {{x_1},{y_1}} \right) = \left( {0,3} \right) in the point slope formula.
Consider the equation (2)
yy1=m(xx1)y - {y_1} = m\left( {x - {x_1}} \right)
Where m=m = \infty , x1=0{x_1} = 0 and y1=3{y_1} = 3 on substitution, we get
y3=(x0)\Rightarrow y - 3 = \infty \left( {x - 0} \right)
Hence, the normal line will be vertical at (0,3)\left( {0,3} \right) and will be of the form x=ax = a .
Since the point of intersection is x=0x = 0 , the normal line will have equation x=0x = 0 . So the yy axis is the normal line at that point.

So, the correct answer is “ x=0x = 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.