JEE Mathematics 2023 Question with Solution

Given:

• Circle $C_1 : x^2 + y^2 - 4x - 2y = \alpha - 5$
• Its mirror image in the line $y = 2x + 1$ is circle $C_2 : 5x^2 + 5y^2 - 10x - 10y + 36 = 0$
• We need $\alpha + r$, where $r$ is the radius of $C_2$

Find: $\alpha + r$

First rewrite the equation of $C_1$:

So,

Hence the center of $C_1$ is $(2,1)$ and its radius is $\sqrt{\alpha}$.

Now rewrite the equation of $C_2$ by dividing by $5$:

Therefore,

Hence the center of $C_2$ is $(1,1)$ and its radius is

The solution computes the distance between the centers:

Then it states:

Therefore,

Therefore, the required value is $2$.

Given: The first circle is $C_1$ and its mirror image in the line $y = 2x + 1$ is $C_2$.

Find: The value of $\alpha + r$.

From

we get

and hence

So the center is $(2,1)$ and radius is $r_1 = \sqrt{\alpha}$.

From

divide by $5$:

Thus,

So the center is $(1,1)$ and radius is $r_2 = \sqrt{\frac{13}{5}}$.

The distance between the centers is

The extracted solution then concludes:

Therefore,

Thus, the final answer is $2$.