A) (3, 1)
B) (1, 3)
C) (3, -1)
D) (-3, 1)
Correct Answer: C
Solution :
\[{{x}^{2}}+{{y}^{2}}-\frac{3}{2}x+\frac{5}{2}y-\frac{7}{2}=0\] \[\Rightarrow {{\left( x-\frac{3}{4} \right)}^{2}}+{{\left( y+\frac{5}{4} \right)}^{2}}-\frac{9}{16}-\frac{25}{16}-\frac{7}{2}=0\] \[\Rightarrow {{\left( x-\frac{3}{4} \right)}^{2}}+{{\left( y+\frac{5}{4} \right)}^{2}}-\frac{45}{8}=0\] 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 \[\equiv \left( \frac{9\times \frac{45}{8}}{\frac{45}{2}},\ \ \frac{1\times \frac{45}{8}}{\frac{45}{2}} \right)\equiv \left( \frac{9}{4},\ \frac{1}{4} \right)\], 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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