Question

Let $S_1=x^2+y^2=9$ and $S_2={(x–2)}^2+y^2=1$. Then the locus of the Centre of a variable circle $S$ which touches $S_1$ internally and $S_2$ externally always passes through the points:

• A. $\left(\frac{1}{2},\pm \frac{\sqrt{5}}{2}\right)$
• B. $\left(2,\pm \frac{3}{2}\right)$
• C. $\left(1,\pm 2\right)$
• D. $(0,\pm \surd 3)$

Answer 1

The correct option is B

$\left(2,\pm \frac{3}{2}\right)$

Explanation for the correct option:

Finding the required points

Step 1: Finding the center and radius of the two circles

The given circle $S_1\equiv x^2+y^2=9$

Therefore, $C_1(0,0)$and $r_1=3$ be the Centre and radius respectively.

Also, $S_2={(x–2)}^2+y^2=1$

And also given $C_2(2,0)$ and , $r_2=1$ be the Centre and radius respectively.

Considering the Centre of the variable circle be $C_3(h,k)$ and the radius be $r$.

Step 2: Illustrating the given information

Step 3: Calculating the distance between the centre

$C_3{C}_1=3–r$

Similarly,

$$\begin{gathered} C_2{C}_3=1+r \\ \therefore C_3{C}_1+C_2{C}_3=4 \end{gathered}$$

Step 4: Using the concept of an ellipse

Therefore, the locus is an ellipse whose foci are $C_1\&C_2$

And major axis is $2a=4$and also $2ae=C_1{C}_2=2$

$$\begin{gathered} \Rightarrow e=\frac{1}{2} \\ \Rightarrow b^2=4\left[1–\frac{1}{4}\right]=3\left[\because e=\sqrt{1-{\left(\frac{b}{a}\right)}^2}\right] \end{gathered}$$

Thus, the center of the ellipse is the midpoint of $C_1\&C_2$is $(1,0)$

Now the equation of the ellipse

$\Rightarrow \frac{{\left(x–1\right)}^2}{2^2}+\frac{y^2}{{\left(\sqrt{3}\right)}^2}=1\left[\because \frac{{\left(x-x_1\right)}^2}{a^2}+\frac{{\left(y-y_1\right)}^2}{b^2}=1\right]$

So now taking $(x,y)=\left(2,\pm \frac{3}{2}\right)$

$$\begin{gathered} \Rightarrow \frac{{\left(2–1\right)}^2}{2^2}+\frac{{\left({\displaystyle \frac{3}{2}}\right)}^2}{{\left(\sqrt{3}\right)}^2}=1 \\ \Rightarrow \frac{1}{4}+\frac{3}{4}=1 \\ \Rightarrow 1=1 \end{gathered}$$

Hence, it always passes through the point $\left(2,\pm \frac{3}{2}\right)$.

Thus, option (B) is the correct answer.