Circle Ques 80
- Consider the family of circles $x^{2}+y^{2}=r^{2}, 2<r<5$. If in the first quadrant, the common tangent to a circle of this family and the ellipse $4 x^{2}+25 y^{2}=100$ meets the coordinate axes at $A$ and $B$, then find the equation of the locus of the mid-points of $A B$.
(1999, 5M)
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Answer:
Correct Answer: 80.$4 x^{2}+25 y^{2}=4 x^{2} y^{2}$
Solution:
Formula:
- Equation of any tangent to circle $x^{2}+y^{2}=r^{2}$ is
$$ x \cos \theta+y \sin \theta=r $$
Suppose Eq. (i) is tangent to $4 x^{2}+25 y^{2}=100$
or $\quad \frac{x^{2}}{25}+\frac{y^{2}}{4}=1$ at $\left(x _1, y _1\right)$
Then, Eq. (i) and $\frac{x x _1}{25}+\frac{y y _1}{4}=1$ are identical
$$ \begin{array}{rlrl} & \therefore & \frac{x _1 / 25}{\cos \theta} & =\frac{\frac{y _1}{4}}{\sin \theta}=\frac{1}{r} \\ \Rightarrow & x _1 & =\frac{25 \cos \theta}{r}, y _1=\frac{4 \sin \theta}{r} \end{array} $$
The line (i) meet the coordinates axes in $A(r \sec \theta, 0)$ and $\beta(0, r \operatorname{cosec} \theta)$. Let $(h, k)$ be mid-point of $A B$.
Then,
$$ h=\frac{r \sec \theta}{2} \quad \text { and } \quad k=\frac{r \operatorname{cosec} \theta}{2} $$
Therefore, $\quad 2 h=\frac{r}{\cos \theta}$ and $\quad 2 k=\frac{r}{\sin \theta}$
$$ \therefore \quad x _1=\frac{25}{2 h} \quad \text { and } \quad y _1=\frac{4}{2 k} $$
As $\left(x _1, y _1\right)$ lies on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{4}=1$, we get
$$ \begin{array}{rlrl} \Rightarrow & \frac{1}{25} \frac{625}{4 h^{2}}+\frac{1}{4} \frac{4}{k^{2}} & =1 \\ \text { or } & \frac{25}{4 h^{2}}+\frac{1}{k^{2}} & =1 \\ 25 k^{2}+4 h^{2} & =4 h^{2} k^{2} \end{array} $$
Therefore, required locus is $4 x^{2}+25 y^{2}=4 x^{2} y^{2}$
BETA
