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  • A circle passes through the origin and has diameter 4 on the

A circle \mathrm{C_{1}} passes through the origin \mathrm{O} and has diameter 4 on the positive \mathrm{x}-axis. The line \mathrm{y=2 x} gives a chord \mathrm{OA} of circle \mathrm{C}_{1}. Let \mathrm{C}_{2} be the circle with \mathrm{OA} as a diameter. If the tangent to \mathrm{C}_{2} at the point \mathrm{A} meets the \mathrm{x}-axis at \mathrm{P} and \mathrm{y}-axis at \mathrm{Q}, then \mathrm{QA} :\mathrm{ AP} is equal to:

Option: 1

1:4


Option: 2

1:5


Option: 3

2:5


Option: 4

1:3


Answers (1)

best_answer

\begin{aligned} &\text{Equation of } \mathrm{C_{1}} \text{ is}\\ &\mathrm{(x-2)^{2}+y^{2}=2^{2}} \\ & \mathrm{\Rightarrow \quad x^{2}+y^{2}-4 x=0}\\ & \text{For point A}\\ & \mathrm{x^{2}+(2 x)^{2}-4 x=0 }\\ & \mathrm{5 x^{2}-4 x=0} \\ & \mathrm{x=\frac{4}{5}} \\ \therefore \mathrm{ A }& \mathrm{\left(\frac{4}{5}, \frac{8}{5}\right)}\\ &\mathrm{Q P} \perp \mathrm{O A}\\ &\therefore Equation of QP\\ &\mathrm{\left(y-\frac{8}{5}\right)=-\frac{1}{2}\left(x-\frac{4}{5}\right)} \\ &\Rightarrow \mathrm{5 y-8=-\frac{1}{2}(5 x-4)} \\ &\Rightarrow \mathrm{10 y-16=-5 x+4 }\\ &\Rightarrow \mathrm{ 5 x+10 y-20=0} \\ &\Rightarrow \mathrm{x+2 y-4=0} \\ &\therefore \mathrm{P(4,0), Q(0,2)}\\ &\text{Let A divide QP in ratio h: 1}\\ & \mathrm{\frac{4 \lambda+0}{R+1}=\frac{4}{5}} \\ \Rightarrow &\mathrm{ 5 \lambda=\lambda+1} \\ \Rightarrow & \mathrm{\lambda=\frac{1}{4}} \\ \therefore & \text { Ratio }=1: 4 \\ \therefore & \text { option (A) } \end{aligned}

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