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Let \mathrm{A} be the point (1,2) and \mathrm{B} be any point onthe curve \mathrm{x}^2+y^2=16. If the centre of the locus of the point \mathrm{P}, which divides the line segment \mathrm{AB} in the ratio 3: 2 is the point \mathrm{C}(\alpha, \beta) then the length of the line segment \mathrm{AC}  is

Option: 1

\frac{6 \sqrt{5}}{5}


Option: 2

\frac{2 \sqrt{5}}{5}


Option: 3

\frac{3 \sqrt{5}}{5}


Option: 4

\frac{4 \sqrt{5}}{5}


Answers (1)

best_answer

\frac{12 \cos \theta+2}{5}=h \Rightarrow 12 \cos \theta=5 \mathrm{~h}-2

sq & add

\begin{aligned} & 144=(5 h-2)^2+(5 k-4)^2 \\ & \left(x-\frac{2}{5}\right)^2+\left(y-\frac{4}{5}\right)^2=\frac{144}{25} \\ & \text { Centre } \equiv\left(\frac{2}{5}, \frac{4}{5}\right) \equiv(\alpha, \beta) \\ & A C=\sqrt{\left(1-\frac{2}{5}\right)^2+\left(2-\frac{4}{5}\right)^2} \\ & =\sqrt{\frac{9}{25}+\frac{36}{25}}=\frac{\sqrt{45}}{5}=\frac{3 \sqrt{5}}{5} \end{aligned}

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shivangi.bhatnagar

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