A) (2, 1)
B) (1, -2)
C) (1, 2)
D) (-1, 2)
Correct Answer: C
Solution :
As the line \[ax+by=0\] touches the circle\[{{x}^{2}}+{{y}^{2}}+2x+4y=0\], distance of the centre (?1, ?2) from the line = radius Þ \[\left| \frac{-a-2b}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|=\sqrt{{{(-1)}^{2}}+{{(-2)}^{2}}}\] Þ \[{{(a+2b)}^{2}}=5({{a}^{2}}+{{b}^{2}})\] Þ \[4{{a}^{2}}-4ab+{{b}^{2}}=0\] Þ \[{{(2a-b)}^{2}}=0\] \ \[b=2a\]. Next, \[ax+by=0\] is a normal to\[{{x}^{2}}+{{y}^{2}}-4x+2y-3=0\], the centre (2, ?1) should lie on \[ax+by=0\] \ \[2a-b=0\Rightarrow b=2a\]. Hence\[a=1\], \[b=2\].You need to login to perform this action.
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