Animated Solution for Mathematics - Circles: If the circles $x^2+y^2+6x+8y+16=0$ and $x^2+y^2+2(3-\sqrt{3})x+2(4-\sqrt{6})y=k+6\sqrt{3}+8\sqrt{6},k>0$, touch internally at the point $P(\alpha ,\beta )$, then $(\alpha +\sqrt{3}{)}^2+(\beta +\sqrt{6}{)}^2$ is equal to ________

Welcome, fellow traveler on the JEE journey. Today, we are not just solving a problem; we are exploring the elegant dance of two circles in the coordinate plane.

Geometry is the language of the universe, and when we see two circles touching internally, we are witnessing a moment of perfect alignment. Let us break down this problem, not as a chore, but as a discovery.

Decoding the Circles

Every circle hides its secrets in its general equation: $x^2+y^2+2gx+2fy+c=0$. Our first task is to bring these circles into the light by completing the square.

For our first circle, $x^2+y^2+6x+8y+16=0$, we group the terms: $(x^2+6x)+(y^2+8y)=-16$. Adding the necessary constants to complete the squares, we get:

This simplifies beautifully to $(x+3{)}^2+(y+4{)}^2=3^2$. We have found our first anchor: center $O_1(-3,-4)$ and radius $r_1=3$.

Now, consider the second circle: $x^2+y^2+2(3-\sqrt{3})x+2(4-\sqrt{6})y=k+6\sqrt{3}+8\sqrt{6}$. By halving the coefficients of $x$ and $y$ and flipping the signs, we find the center $O_2(-3+\sqrt{3},-4+\sqrt{6})$.

The radius $r_2$ is found using $r=\sqrt{g^2+f^2-c}$. After substituting and simplifying, the complex terms involving $\sqrt{3}$ and $\sqrt{6}$ vanish, leaving us with:

The Geometric Dance

Here is the core of our problem. The circles touch internally, meaning the smaller circle is nestled inside the larger one.

For them to kiss at exactly one point, the distance between their centers, $O_1{O}_2$, must be exactly the difference of their radii:

Let us calculate the distance $O_1{O}_2$ using the distance formula:

The constants $-3$ and $-4$ cancel out, leaving us with $\sqrt{(\sqrt{3}{)}^2+(\sqrt{6}{)}^2}=\sqrt{3+6}=3$. The distance is exactly $3$.

Since $r_1=3$ and the condition is $O_1{O}_2=|r_1-r_2|$, we have $3=|3-r_2|$. Given $k>0$, we find $r_2=6$, which implies $\sqrt{k+34}=6$, so $k=2$.

The Point of Contact

Now, where do they touch? The point of contact $P(\alpha ,\beta )$ lies on the line connecting the centers.

Because the circles touch internally, $P$ divides the segment $O_1{O}_2$ externally in the ratio of their radii, $r_1:r_2=3:6=1:2$.

Using the external division formula $P=\frac{m{x}_2-n{x}_1}{m-n}$, we find:

Similarly, for $\beta$, we calculate:

We are asked to evaluate $(\alpha +\sqrt{3}{)}^2+(\beta +\sqrt{6}{)}^2$. Substituting our values:

Look at that! The square roots that once seemed so daunting have completely vanished, leaving us with a clean, perfect integer.

The final answer is 25. This is the beauty of mathematics—no matter how complex the path, the truth is often simple and elegant.