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  • If the radius of a circle which passes through the point (2,0) and whose centre is the limit of the point of intersection of the lines <img alt="3 x+5 y=1" src="https://entrancecorner.onc

If the radius of a circle which passes through the point (2,0) and whose centre is the limit of the point of intersection of the lines 3 x+5 y=1 and (2+c) x+5 c^2 y=1 as c \rightarrow 1 \text { is } \frac{\sqrt{\lambda}}{25} , then the value of \lambda must be 

Option: 1

1200


Option: 2

1600


Option: 3

1225


Option: 4

1601


Answers (1)

best_answer

Solving the equations

\begin{array}{r} (2+c) x+5 c^2 y=1 \\ \text{and}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3 x+5 y=1 \end{array}

\\(2+c) x+5 c^2\left(\frac{1-3 x}{5}\right)=1\\ \\\text{then} \: \: \: \: \: \: \: \: \: \: \: \: \: \quad(2+c) x+c^2(1-3 x)=1

\begin{array}{ll} \therefore & x=\frac{1-c^2}{2+c-3 c^2} \\ \\\text { or } & x=\frac{(1+c)(1-c)}{(3 c+2)(1-c)}=\frac{1+c}{3 c+2} \end{array}

\begin{aligned} \therefore &\: \: \: \: \: \: \: \: \: \: \: \: \: =\lim _{c \rightarrow 1} \frac{1+c}{3 c+2} \\ & \: \: \: \: \: \: \: \: \: \: \: x=\frac{2}{5} \end{aligned}

\therefore        y=\frac{1-3 x}{5}=\frac{1-\frac{6}{5}}{5}=-\frac{1}{25}

Therefore the centre of the required circle is \left(\frac{2}{5},-\frac{1}{25}\right) but circle passes through (2,0)

\therefore Radius of the required circle =\sqrt{\left(\frac{2}{5}-2\right)^2+\left(-\frac{1}{25}-0\right)^2}

\begin{aligned} & =\sqrt{\frac{64}{25}+\frac{1}{625}}=\sqrt{\frac{1601}{625}}=\frac{\sqrt{1601}}{25}=\frac{\sqrt{\lambda}}{25} \\ \\\therefore \quad \lambda & =1601 \end{aligned}

Posted by

Sanket Gandhi

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