JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let the tangents drawn from the origin to the circle, \(x^{2}+y^{2}-8 x-4 y+16=0\) touch it at the points \(A\) and \(B .\) The \((A B)^{2}\) is equal to
- A \(\frac{52}{5}\)
- B \(\frac{32}{5}\)
- C \(\frac{56}{5}\)
- D \(\frac{64}{5}\)
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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