The equation of the chord of the circle x²+y²=81 which is bisected at the point (-2,3) is

Correct Answer is : 2x−3y=−132x-3y=-132x−3y=−13

Correct Answer is : 2x−3y=−132x-3y=-132x−3y=−13

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KEAM 2010 Math Paper

Question : 119 of 120

Marks: +1, -0

x^{2}+y^{2}=81

which is bisected at the point (-2,3) is

Solution:

The equation of line joining the centre of circle

x^{2}+y^{2}=81

and the point

(-2,3)

y-0=\; \frac{3-0}{-2-0}(x-0)

\Rightarrow y=-\; \frac{3}{2} x

\Rightarrow 3x+2y=0

...(i)

The equation of line perpendicular to line (i) is

2x-3y+\lambda=0

\because

It is possess through

(-2,3)

\therefore 2 \cdot (-2)-3 \cdot (3)+\lambda=0

\Rightarrow \lambda=13

\therefore

Required equation of line is

2x-3y+13=0

\Rightarrow 2x-3y=-13

Alternate

Required equation of line is

S_{1}=T

(-2)^{2}+(3)^{2}-81=-2x_{1}+3y_{1}-81

\Rightarrow 4+9=-2x_{1}+3y_{1}

\Rightarrow 2x_{1}-3y_{1}=-13

Replace

x_{1}

and

y_{1}

x

and

y

respectively,

2x-3y=-13

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