A) a tangent
B) cut at two real points
C) cut at two imaginary points
D) None of the above
Correct Answer: A
Solution :
Equation of polar of the circle\[{{x}^{2}}+{{y}^{2}}=5\]with respect to the point\[(1,-2)\]is \[x(1)+y(-2)=5\] \[\Rightarrow \] \[x-2y=5\] \[\Rightarrow \] \[x-2y-5=0\] Centre of the circle \[{{x}^{2}}\text{+}{{y}^{2}}-8x+6y+20=0\]is\[(4,-3)\]and radius\[=\sqrt{16+9-20}=\sqrt{5}\] Perpendicular distance from the centre\[(4,-3)\] to the line\[x-2y-5=0\] \[=\left| \frac{4+6-5}{\sqrt{1+4}} \right|\] \[=\left| \frac{5}{\sqrt{5}} \right|=\sqrt{5}=\]radius of circle Hence, this is the tangent to the circle.
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