Circle And System Of Circles Question 101
Question: The pole of the straight line $ 9x+y-28=0 $ with respect to circle $ 2x^{2}+2y^{2}-3x+5y-7=0 $ , is
[RPET 1990, 99; MNR 1984; UPSEAT 2000]
Options:
A) (3, 1)
B) (1, 3)
C) (3, -1)
D) (-3, 1)
Show Answer
Answer:
Correct Answer: C
Solution:
$ x^{2}+y^{2}-\frac{3}{2}x+\frac{5}{2}y-\frac{7}{2}=0 $
$ \Rightarrow {{( x-\frac{3}{4} )}^{2}}+{{( y+\frac{5}{4} )}^{2}}-\frac{9}{16}-\frac{25}{16}-\frac{7}{2}=0 $
$ \Rightarrow {{( x-\frac{3}{4} )}^{2}}+{{( y+\frac{5}{4} )}^{2}}-\frac{45}{8}=0 $ Put $ X=x-\frac{3}{4} $ and $ Y=y+\frac{5}{4} $ , we get the equation of circle $ X^{2}+Y^{2}-\frac{45}{8}=0 $ and the line $ 9X+Y-\frac{45}{2}=0 $
Hence pole $ \equiv ( \frac{9\times \frac{45}{8}}{\frac{45}{2}},\ \ \frac{1\times \frac{45}{8}}{\frac{45}{2}} )\equiv ( \frac{9}{4},\ \frac{1}{4} ) $ , But $ x=\frac{9}{4}+\frac{3}{4} $ and $ y=\frac{1}{4}-\frac{5}{4}=-1 $ .
Hence the pole is (3, -1).
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