Question

Find the equation of normal to the circle x²+y²-4x-6y+11=0 at (3,2) also find the other point where the normal meets the circl​

Answer 1

Answer:

First circle - solve by completing the square:

x²+ y² - 4x - 6y - 12 = 0

(x² - 4x) + (y² - 6y) - 12 = 0

(x² - 4x + 4) + (y² - 6y + 9) - 25 = 0

(x-2)² + (y-3)² = 25

So this circle has its center at the point (2,3) and radius 5.

Do the same for the second circle:

x² + y² + 6x + 18y + 26 = 0

(x² + 6x) + (y² + 18y) + 26 = 0

(x² + 6x + 9) + (y² + 18y + 81) - 64 = 0

(x+3)² + (y+9)² = 64

So this circle has its center at the point (-3, -9) and radius 8.

How do we know they touch each other? The x coordinates differ by 5, the y coordinates differ by 12, and the sum of the two radii is 13, and 5/12/13 is a Pythagorean triplet. So the radii of the two circles form the hypotenuse of a right triangle, like this:

The point of tangency should be (+1/13, -21/13.) Since the slope of the line that connects the two radii is 12/5, the slope of the tangent line must be -5/12.

I hope it will be help you

Answer 2

The equation of normal to the circle x² + y² - 4x - 6y + 11 = 0 at (3,2)  is  x + y - 5 = 0.

The other point where the normal meets the circle is ( 1, 4 ).

Given: The equation of the circle x² + y² - 4x - 6y + 11 = 0, point (3, 2).

To Find: The equation of normal to the circle and the other point where the normal meets the circle.

Solution:

We are given the equation of circle as,

             S = x² + y² - 4x - 6y + 11 = 0

Now, center (C) = ( - g, - f ) = ( 2, 3 )

We know that equation of a tangent of a circle can be found using the,

            S1 = 0

   ⇒ xx1 + yy1 + g ( x + x1 ) + f ( y + y1 ) + c = 0      where  ( x1, y1 ) = ( 3, 2 )

       ⇒ 3x + 2y - 2 × ( x + 3 ) - 3 × ( y + 2 ) + 11 = 0

       ⇒ 3x + 2y - 2x - 6 - 3y - 6 + 11 = 0

       ⇒ x - y - 1 = 0

We know that the slope of normal ( N.S ) is perpendicular to the slope of tangent ( T.S ),

   So, the slope of the tangent = 1

   ∴ slope of normal = - 1 / T.S = - 1 / 1

                                 = - 1

So, the equation of normal at (3, 2) is given by

y - y1 = N.S ( x - x1 )     [ where, N.S = slope of normal, ( x1, y1 ) = ( 3, 2 )]

  ⇒ y - 2 = - 1 ( x - 3 )

  ⇒  y - 2 = - x + 3

  ⇒ x + y - 5 = 0

Hence, equation of normal to the circle is x + y - 5 = 0.

Now, we need to find the other point where the normal meets the circle

Let other point where the normal meets are ( x2, y2 ).

We know that the center ( 2, 3 ) divides the normal into two equal parts at ( 3, 2 ) and ( x2, y2 ), thus;

      ( 2, 3 ) ≡ (( 3 + x2) /2 , ( 2 + y2)/2 )

  ⇒ ( 3 + x2) /2 = 2 ⇒ x2 = 1

  ⇒ ( 2 + y2)/2 = 3  ⇒ y2 = 4

The other point where the normal meets the circle is ( 1, 4 ).

Hence, the equation of normal to the circle x² + y² - 4x - 6y + 11 = 0 at (3,2)  is  x + y - 5 = 0.

The other point where the normal meets the circle is ( 1, 4 ).

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