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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

माना मूल बिन्दु से वृत्त \(x^{2}+y^{2}-8 x-4 y+16=0\) पर खींची गई स्पर्श रेखायें इसे बिन्दुओं \(A\) तथा \(B\) पर स्पर्श करती है। तो \(( AB )^{2}\) बराबर है 

  1. A \(\frac{52}{5}\)
  2. B \(\frac{32}{5}\)
  3. C \(\frac{56}{5}\)
  4. D \(\frac{64}{5}\)
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Answer & Solution

Correct Answer

(D) \(\frac{64}{5}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{R}=\sqrt{16+4-16}=2\) \(\mathrm{L}=\sqrt{\mathrm{S}_{1}}=4\) \(\mathrm{AB}(\text { Chord of contact })=\frac{2 \mathrm{LR}}{\sqrt{\mathrm{L}^{2}+\mathrm{R}^{2}}}=\frac{8}{\sqrt{5}}\) \((\mathrm{AB})^{2}=\frac{64}{5}\)
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