JEE Mathematics 2024 Question with Solution

Given: The circles have centers $(a,3)$ and $(8,1)$ from the question, and they touch externally at $(6,6)$. The solution working uses the intended center of the second circle as $\left(8,\frac{15}{2}\right)$ and evaluates the required expression as $(\alpha + \beta) + 4(r_1^2 + r_2^2)$.

Find: The value of the required expression and hence the correct option.

Using the section formula for internal division in the ratio $2:1$:

$$\frac{16 + \alpha}{3} = 6$$ $$\frac{15 + \beta}{3} = 6$$

So,

$$\alpha = 2, \qquad \beta = 3$$

Hence the center of $C_1$ is $(2,3)$.

Now the distance between the centers is

$$\sqrt{(2-8)^2 + \left(3-\frac{15}{2}\right)^2}$$ $$= \sqrt{36 + \frac{81}{4}} = \sqrt{\frac{225}{4}} = \frac{15}{2}$$

Since the circles touch externally,

$$r_1 + r_2 = \frac{15}{2}$$

Also, the common point $(6,6)$ lies on $C_1$, so

$$(6-2)^2 + (6-3)^2 = r_1^2$$ $$16 + 9 = 25$$

Thus,

$$r_1 = 5$$

Therefore,

$$r_2 = \frac{15}{2} - 5 = \frac{5}{2}$$

Now,

$$\alpha + \beta = 2 + 3 = 5$$

and

$$r_1^2 + r_2^2 = 25 + \left(\frac{5}{2}\right)^2 = 25 + \frac{25}{4} = \frac{125}{4}$$

Hence,

$$4(r_1^2 + r_2^2) = 125$$

Therefore,

$$(\alpha + \beta) + 4(r_1^2 + r_2^2) = 5 + 125 = 130$$

So the correct option is B.

Note: The given question text and the first solution contain notation inconsistencies, but the second solution clearly concludes the answer as $130$.

Given: The touching point is $(6,6)$ and the point divides the line joining the centers internally in the ratio $2:1$.

Find: The final numerical value.

From the solution, let the centers be $(\alpha,\beta)$ and $\left(8,\frac{15}{2}\right)$. By section formula,

$$6 = \frac{2\cdot 8 + \alpha}{3}$$ $$6 = \frac{2\cdot \frac{15}{2} + \beta}{3}$$

This gives

$$\alpha = 2, \qquad \beta = 3$$

So one center is $(2,3)$.

Because $(6,6)$ is the point of contact, it lies on the first circle. Hence its radius is the distance from $(2,3)$ to $(6,6)$:

$$r_1 = \sqrt{(6-2)^2 + (6-3)^2}$$ $$= \sqrt{16+9} = 5$$

The distance between centers is

$$\sqrt{(8-2)^2 + \left(\frac{15}{2}-3\right)^2} = \frac{15}{2}$$

For external touching,

$$r_1 + r_2 = \frac{15}{2}$$

Therefore,

$$r_2 = \frac{15}{2} - 5 = \frac{5}{2}$$

Now compute

$$(\alpha + \beta) + 4(r_1^2 + r_2^2)$$ $$= (2+3) + 4\left(25 + \frac{25}{4}\right)$$ $$= 5 + 4\cdot \frac{125}{4}$$ $$= 5 + 125 = 130$$

Therefore, the required value is $130$, so the correct option is B.