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Consider the family of circles \mathrm{x^2+y^2=r^2, 2<r<5}. If in the first quadrant, the common tangent to a circle of this family and the ellipse \mathrm{4 x^2+25 y^2=100} meets the coordinate axes at A and B, then the equation of the locus of the mid-point of AB is,

Option: 1

\mathrm{4 x^2+25 y^2=4 x^2 y^2}


Option: 2

\mathrm{25 x^2+4 y^2=2 x^2 y^2}


Option: 3

\mathrm{4 x^2+25 y^2=x^2 y^2}


Option: 4

\mathrm{2 x^2+25 y^2=x^2 y^2}


Answers (1)

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\mathrm{\text { Equation of any tangent to circle } x^2+y^2=r^2 \text { is }}

\mathrm{x \cos \theta+y \sin \theta=r}          ...[1]

\mathrm{\text { Let equation (1) is tangent to } 4 x^2+25 y^2=100}

\mathrm{\text { Then the equation (1) and } \frac{x x_1}{25}+\frac{y y_1}{4}=1}     ...[2]

are identical

\mathrm{\therefore \frac{x_1 / 25}{\cos \theta}=\frac{y_1 / 4}{\sin \theta}=\frac{1}{r} \Rightarrow x_1=\frac{25 \cos \theta}{r}, y_1=\frac{4 \sin \theta}{r}}

The line (1) meets the coordinate axes at A(r secθ, 0) and
B(0, r cosecθ).
Let (h, k) be mid point of AB.

\mathrm{\begin{aligned} & \therefore \quad h=\frac{r \sec \theta}{2} \text { and } k=\frac{r \operatorname{cosec} \theta}{2} \\ & \therefore \quad 2 h=\frac{r}{\cos \theta}, 2 k=\frac{r}{\sin \theta} \\ & \therefore \quad x_1=\frac{25}{2 h} \text { and } y_1=\frac{4}{2 k} \end{aligned}}

\mathrm{\text { Since }\left(x_1, y_1\right) \text { lies on the ellipse } \frac{x^2}{25}+\frac{y^2}{4}=1}

\mathrm{\text { Therefore, } \frac{1}{25}\left(\frac{625}{4 h^2}\right)+\frac{1}{4}\left(\frac{4}{k^2}\right)=1}

\mathrm{\Rightarrow \frac{25}{4 h^2}+\frac{1}{k^2}=1 \text { or } 25 k^2+4 h^2=4 h^2 k^2}

\mathrm{\text { Hence required equation of locus is } 4 x^2+25 y^2=4 x^2 y^2}

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