Home
Class 12
MATHS
Find the equation of the normal to the c...

Find the equation of the normal to the circle circle `x^(2)+y^(2)-5x+2y-48=0` at the point (5,6).

A

`2x-y-7=0, 2x+y-9=0`

B

`2x+y-7=0, 2x+y+9=0`

C

`2x+y+7=0, 2x+y+9=0`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
If ordinate is - 1 then abscissa may be 3 or 5. Hence normal are lines passing through (4, 1) and (3, -1) and other passes through (4, 1) (5, -1)
Promotional Banner

Topper's Solved these Questions

  • JEE MAIN - 5

    VMC MODULES ENGLISH|Exercise PART III : MATHEMATICS (SECTION-2)|5 Videos

  • INVERSE TRIGONOMETRY

    VMC MODULES ENGLISH|Exercise JEE ADVANCED ( ARCHIVE )|10 Videos

  • JEE MAIN REVISION TEST - 30 | JEE -2020

    VMC MODULES ENGLISH|Exercise MATHEMATICS|25 Videos

Similar Questions

Explore conceptually related problems

The equation of the normal to the circle x^(2)+y^(2)+6x+4y-3=0 at (1,-2) to is

Find the equation of the normals to the circle x^2+y^2-8x-2y+12=0 at the point whose ordinate is -1

Find the equation of the normals to the circle x^2+y^2-8x-2y+12=0 at the point whose ordinate is -1

Find the equation of the normal to the circle x^2+y^2-2x=0 parallel to the line x+2y=3.

Find the equation of the normal to the circle x^2+y^2-2x=0 parallel to the line x+2y=3.

Find the equation of the normal to the circle x^(2) + y^(2)- 4x-6y+11=0 at (3,2) . Also find the other point where the normal meets the circle.

Find the equation of the normal to the circle x^2+y^2=9 at the point (1/(sqrt(2)),1/(sqrt(2))) .

Find the equation of the tangent to the circle x^2 + y^2-30x+6y+109=0 at (4, -1)

Find the equation of tangent and normal at (3, 2) of the circle x^(2) + y^(2) - 3y - 4 = 0.

The equation of the chord of the circle x^(2)+y^(2)-6x+8y=0 which is bisected at the point (5, -3), is