TS EAMCET · Maths · Circle
The pole of the line \(x-5 y-7=0\) with respect to the circle \(S \equiv x^2+y^2-2 x+4 y+1=0\) is \(P(a, b)\). If \(C\) is the centre of the circle \(S=0\) then \(P C=\)
- A \(\sqrt{a+b-1}\)
- B \(\sqrt{a^2+b^2-1}\)
- C \(\sqrt{a^3+b^3-1}\)
- D \(3 a b\)
Answer & Solution
Correct Answer
(C) \(\sqrt{a^3+b^3-1}\)
Step-by-step Solution
Detailed explanation
\(S \equiv x^2+y^2-2 x+4 y+1=0 \Rightarrow x-5 y-7=0\) Pole is given by \(\frac{a-1}{1}=\frac{b+2}{-5}=\frac{-a+2 b+1}{-7}\) \(\begin{aligned} & \Rightarrow(a, b)=(0,3) \\ & \text { Centre }(C)=(1,-2) \\ & \therefore P C=\sqrt{26}=\sqrt{a^3+b^3-1} . \end{aligned}\)
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