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The locus of the midpoints of the chord of the circle. \mathrm{x^2+y^2=25} which is tangent to the hyperbola, \mathrm{\frac{x^2}{9}-\frac{y^2}{16}=1} is
 

Option: 1

\mathrm{ \left(x^2+y^2\right)^2-9 x^2-16 y^2=0}


Option: 2

\mathrm{ \left(x^2+y^2\right)^2-9 x^2+144 y^2=0}


Option: 3

\mathrm{ \left(x^2+y^2\right)^2-16 x^2+9 y^2=0}


Option: 4

\mathrm{ \left(x^2+y^2\right)^2-9 x^2+16 y^2=0}


Answers (1)

best_answer

Let (h, k) be the mid-point of the chord of circle \mathrm{x^2+y^2=25} with centre (0,0).
\therefore Equation of chord is

\mathrm{h x+k y=h^2+k^2 \Rightarrow y=-\frac{h}{k} x+\frac{h^2+k^2}{k} \quad \quad \dots(i)}

Now, (i) will be tangent to hyperbola \mathrm{\frac{x^2}{9}-\frac{y^2}{16}=1 } if

\begin{aligned} &\mathrm{ c^2=a^2 m^2-b^2 }\\ \Rightarrow &\mathrm{ \left(\frac{h^2+k^2}{k}\right)^2=9\left(\frac{-h}{k}\right)^2-(16) \quad \text { [Using (i)] } }\\ \Rightarrow &\mathrm{ \left(h^2+k^2\right)^2=9 h^2-16 k^2 }\\ \therefore &\mathrm{ \text { Required locus is }\left(x^2+y^2\right)^2-9 x^2+16 y^2=0}\\ \end{aligned}

Posted by

Irshad Anwar

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