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  • Let  and . Then the locus of centre of a variable cir

Let S_1:x^2 +y^2=9 and S_2 : (x-2)^2 +y^2=1. Then the locus of centre of a variable circle S which touches S_1 internally S_2 externally always passes through the points:
Option: 1 (0, \pm \sqrt{3})
Option: 2 (1, \pm 2)
Option: 3 (\frac{1}{2}, \pm \frac{\sqrt{5}}{2})  
Option: 4 (2, \pm \frac{3}{2})

Answers (1)

best_answer

\\\mathrm{S}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}=9\\r_1=3,\;\;A(0,0)

\\S_{2}:(x-2)^{2}+y^{2}=1\\r_2=1,\;B(2,0)

\because \mathrm{c}_{1} \mathrm{c}_{2}=\mathrm{r}_{1}-\mathrm{r}_{2}

therefore given circle are touching internally Let a veriable circle with P and radius
r

\\\Rightarrow \mathrm{PA}=\mathrm{r}_{1}-\mathrm{r} \text { and } \mathrm{PB}=\mathrm{r}_{2}+\mathrm{r} \\ \Rightarrow \mathrm{PA}+\mathrm{PB}=\mathrm{r}_{1}+\mathrm{r}_{2} \\ \Rightarrow \mathrm{PA}+\mathrm{PB}=4 \quad(>\mathrm{AB})

⇒ Locus of P is an ellipse with foci at A(0,0) and B(2,0) and length of major axis is 2a = 4

e = 1/2

\Rightarrow \text{centre is at (1,0) and } b^{2}=a^{2}\left(1-e^{2}\right)=3

\begin{aligned} &\text { Equation of Ellipse is } \frac{(\mathrm{x}-1)^{2}}{2^{2}}+\frac{\mathrm{y}^{2}}{3}=1 \quad \Rightarrow \mathrm{e}=\frac{1}{2}\\ &\Rightarrow \mathrm{b}^{2}=3\\ &\left(2, \pm \frac{3}{2}\right) \text { satisfied it } \end{aligned}

Posted by

himanshu.meshram

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