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  • A circle of constant radius passes through the origin <img alt="\mathrm{O}" src="https://entrancecorn

A circle of constant radius \mathrm{r} passes through the origin \mathrm{O}, and cuts the axes at \mathrm{A}$ and $\mathrm{B}. The locus of the foot of the perpendicular from \mathrm{O\: to\, A B} is \mathrm{ \left(x^2+y^2\right)\left(x^{-2}+y^{-2}\right)=k^2 \text { where } k=}
 

Option: 1

\mathrm{4r}


Option: 2

\mathrm{3r}


Option: 3

\mathrm{2r}


Option: 4

\mathrm{r}


Answers (1)

best_answer

Let the coordinates of \mathrm{A}$ and $\mathrm{B} are \mathrm{(\mathrm{a}, 0) and (0, \mathrm{~b})

\therefore Equation of \mathrm{AB} is
\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1                ..............(i)
center of circle lie on line \mathrm{A B}, since \mathrm{A B} is

diameter of the circle
(\because \angle \mathrm{AOB}=\pi / 2)
\therefore \quad co-ordinate of center \mathrm{C} is \mathrm{C} \equiv\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)

since the radius of circle =\mathrm{r}
\therefore \mathrm{r}=\mathrm{AC}=\mathrm{CB}=\mathrm{OC}=\sqrt{\left(0-\frac{\mathrm{a}}{2}\right)^2+\left(0-\frac{\mathrm{b}}{2}\right)^2}

\begin{aligned} & =\sqrt{\frac{a^2+b^2}{4}} \\ \end{aligned}

\mathrm{\therefore \quad a^2+b^2 =4 r^2}             ...............(ii)
Equation of \mathrm{OM} which is perpendicular to \mathrm{AB} is
\text { ax }- \text { by }=\lambda
It passes through (0,0)

\mathrm{\therefore 0=\lambda }

\therefore Equation of \mathrm{OM} is  \mathrm{ax}-\mathrm{by}=0   .............(iii)
Solving (i) and (iii), we get

\mathrm{ a=\frac{x^2+y^2}{x} \text { and } b=\frac{x^2+y^2}{y} }

Substituting the values of a and b in (ii), we get

   \mathrm{ \left(x^2+y^2\right)^2\left(\frac{1}{x^2}+\frac{1}{y^2}\right)=4 r^2 }
or \mathrm{ \left(x^2+y^2\right)^2\left(x^{-2}+y^{-2}\right)=4 r^2}
which is the required locus.

Hence option 3 is correct

Posted by

Ritika Jonwal

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