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  • The pole of the straight line with respect to the circle <img alt="\ma

The pole of the straight line \mathrm{9 x+y-28=0} with respect to the circle \mathrm{2 x^{2}+2 y^{2}-3 x+5 y-7=0} is

Option: 1

(3,1)


Option: 2

(1,3)


Option: 3

(3,-1)


Option: 4

(-3,1)


Answers (1)

Let \mathrm{(h, k)} be the pole of \mathrm{2 x^{2}+2 y^{2}-3 x+5 y-7=0}

\mathrm{\Rightarrow} The polar is
2 \mathrm{hx}+2 \mathrm{ky}-\frac{3}{2}(\mathrm{x}+\mathrm{h})+\frac{5}{2}(\mathrm{y}+\mathrm{k})-7=0
4 \mathrm{hx}+4 \mathrm{ky}-3(\mathrm{x}+\mathrm{h})+5(\mathrm{y}+\mathrm{k})-14=0

\Rightarrow \quad(4 h-3) x+(4 k+5) y-(3 h-5 k+14)=0

Comparing with 9 \mathrm{x}+\mathrm{y}-28=0
4 \mathrm{~h}-3=9 \quad \Rightarrow \quad \mathrm{h}=3$ and $4 \mathrm{k}+5=1 \quad \Rightarrow \quad \mathrm{k}=-1

and 3 \mathrm{~h}-5 \mathrm{k}+14=28$ put $\mathrm{h}=3$ and $\mathrm{k}=-1
9+5+14=28=\text { R.H.S. }
\Rightarrow \quad(3,-1) is the pole.

Hence (C) is the correct answer.

Posted by

Ramraj Saini

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