The equation of the normal to the circle x²+y²+6x+4y-3=0 at (1,-2) is

Correct Answer is : y+2=0y+2=0y+2=0

Correct Answer is : y+2=0y+2=0y+2=0

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EAMCET Engineering 2001 Solved Paper

Section: Mathematics

Question : 126 of 200

Marks: +1, -0

x^2+y^2+6x+4y-3=0

(1,-2)

Solution:

Given circle is

x^2+y^2+6x+4y-3=0

Equation of the tangent at

(x_1, y_1)

x x_1+y y_1+3(x+x_1)+2(y+y_1)-3=0

If this line passes through

(1,-2)

, then

\;x-2y+3(x+1)+2(y-2)-3=0

\;\Rightarrow\;\; x-2y+3x+3+2y-4-3=0

\;\Rightarrow\;\; 4x-4=0

\;\Rightarrow\;\; x-1=0

Equation of the normal is

0\cdot x-1\cdot y=\lambda

It passes through

(1,-2)

-(-2)\;=\lambda \Rightarrow \lambda=2

\Rightarrow \;\; -y\;=2 \Rightarrow y+2=0

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