JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let a variable line passing through the centre of the circle \(x^2+y^2-16 x-4 y=0\), meet the positive co-ordinate axes at the point \(\mathrm{A}\) and \(\mathrm{B}\). Then the minimum value of \(\mathrm{OA}+\mathrm{OB}\), where \(\mathrm{O}\) is the origin, is equal to
- A \(12\)
- B \(18\)
- C \(20\)
- D \(24\)
Answer & Solution
Correct Answer
(B) \(18\)
Step-by-step Solution
Detailed explanation
\((y-2)=m(x-8)\) \(\Rightarrow x \text {-intercept }\) \(\Rightarrow\left(\frac{-2}{m}+8\right)\) \(\Rightarrow y \text {-intercept }\) \(\Rightarrow(-8 \mathrm{~m}+2)\) \(\Rightarrow \mathrm{OA}+\mathrm{OB}=\frac{-2}{\mathrm{~m}}+8-8 \mathrm{~m}+2\)…
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