Exams Joint Entrance Examination Mathematics Circle Question 47 Question 47 of 219 Medium MCQ
Let a variable line passing through the centre of the circle x 2 + y 2 − 16 x − 4 y = 0 x^2+y^2-16 x-4 y=0 x 2 + y 2 − 16 x − 4 y = 0 , meet the positive co-ordinate axes at the points A A A and B B B . Then the minimum value of O A + O B O A+O B O A + O B , where O O O is the origin, is equal to
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\[ ( y − 2 ) = m ( x − 8 ) ⇒ x -intercept ⇒ ( − 2 m + 8 ) ⇒ y -intercept ⇒ ( − 8 m + 2 ) ⇒ O A + O B = − 2 m + 8 − 8 m + 2 f ′ ( m ) = 2 m 2 − 8 = 0 ⇒ m 2 = 1 4 ⇒ m = − 1 2 ⇒ f ( − 1 2 ) = 18 ⇒ Minimum = 18 \begin{aligned}
& (y-2)=m(x-8) \\
& \Rightarrow x \text {-intercept } \\
& \Rightarrow\left(\frac{-2}{m}+8\right) \\
& \Rightarrow y \text {-intercept } \\
& \Rightarrow(-8 \mathrm{~m}+2) \\
& \Rightarrow \mathrm{OA}+\mathrm{OB}=\frac{-2}{\mathrm{~m}}+8-8 \mathrm{~m}+2 \\
& \mathrm{f}^{\prime}(\mathrm{m})=\frac{2}{\mathrm{~m}^2}-8=0 \\
& \Rightarrow \mathrm{m}^2=\frac{1}{4} \\
& \Rightarrow \mathrm{m}=\frac{-1}{2} \\
& \Rightarrow \mathrm{f}\left(\frac{-1}{2}\right)=18 \\
& \Rightarrow \text { Minimum }=18
\end{aligned} ( y − 2 ) = m ( x − 8 ) ⇒ x -intercept ⇒ ( m − 2 + 8 ) ⇒ y -intercept ⇒ ( − 8 m + 2 ) ⇒ OA + OB = m − 2 + 8 − 8 m + 2 f ′ ( m ) = m 2 2 − 8 = 0 ⇒ m 2 = 4 1 ⇒ m = 2 − 1 ⇒ f ( 2 − 1 ) = 18 ⇒ Minimum = 18 \] Answer the question to unlock