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  • The locus of the mid point of a chord of the circle  which subtend

The locus of the mid point of a chord of the circle \mathrm{x^2+y^2=4} which subtends a right angle at the origin is

Option: 1

\mathrm{x+y=2}


Option: 2

\mathrm{x^2+y^2=1}


Option: 3

\mathrm{x^2+y^2=2}


Option: 4

\mathrm{x+y=1}


Answers (1)

Let mid point be (h, k), then equation of chord is

\mathrm{h x+k y=h^2+k^2}

Now on homoginising  \mathrm{x^2+y^2=4} with the help of this chord 

\mathrm{x^2+y^2=4\left(\frac{h x+k y}{h^2+k^2}\right)^2}

Now as subtended angle is \mathrm{90^{\circ}}

Coeff. of \mathrm{x^2+} Coeff. of \mathrm{y^2=0}

\mathrm{ 1-\frac{4 h^2}{\left(h^2+k^2\right)}+1-\frac{4 k^2}{\left(h^2+k^2\right)}=0 }

\mathrm{\left(\begin{array}{c} \text { represents a pair of straight lines passing through } \\ \text { origin and intersection points of circle and chords } \end{array}\right)}

\mathrm{\begin{aligned} & 2\left(x^2+y^2\right)^2=4\left(x^2+y^2\right) \\ & x^2+y^2=2 \end{aligned}}

Posted by

Sumit Saini

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