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  • Find the equation of the normal to the circle <img alt="x^{2}+y^{2}+ 2 x+4 y-5=0" src="https://entrancecorner.oncodecogs.com/gif.latex?x%5E%7B2%7D&plus;y%5E%7B2%7D&plus;%202%20x&plus;4%

Find the equation of the normal to the circle x^{2}+y^{2}+ 2 x+4 y-5=0 at (-2,1)

Option: 1

3x+y+5=0


Option: 2

x+5y+9=0


Option: 3

5x-y+9=0


Option: 4

4x+y+7=0


Answers (1)

best_answer

The given circle is x^{2}+y^{2}+ 2 x+4 y-5=0 

Its centre is (-1, -2)

So the normal passes through (-2,1) and (-1,-2)

Slope of this line is -3

Equation of normal

(y + 2) = -3 (x+ 1)

3x + y + 5 = 0 

 

Posted by

Riya

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