JEE Mathematics 2023 Question with Solution

Given: The circle is

$$x^2 + y^2 - 2x = 0$$

and the line $ax + by = 0$ cuts it at $A(\alpha,0)$ and $B(1,\beta)$.

Find: The image of the circle having $AB$ as diameter in the line $x+y+2=0$.

From the given coordinates of the intersection points, the only possibility is

$$\alpha = 0, \quad \beta = 1$$

So the endpoints of the diameter are $A(0,0)$ and $B(1,1)$.

Hence, the circle with $AB$ as diameter has equation

$$x^2 + y^2 - x - y = 0$$

Its image in the line $x+y+2=0$ is the circle

$$x^2 + y^2 + 5x + 5y + 12 = 0$$

Therefore, the correct option is A.

Note: The solution also contains a contradictory label saying option B, but the worked result is $x^2 + y^2 + 5x + 5y + 12 = 0$, which matches option A.

Given: The circle with diameter $AB$ has already been obtained as

$$x^2 + y^2 - x - y = 0$$

Find: Its reflected image in the line $x+y+2=0$.

Write the circle in center-radius form. Its center is the midpoint of $A(0,0)$ and $B(1,1)$, namely

$$\left(\frac{1}{2},\frac{1}{2}\right)$$

The radius remains unchanged under reflection.

Reflect the center across the line

$$x+y+2=0$$

Using the reflection formula for a point $\left(x_1,y_1\right)$ in the line $ax+by+c=0$,

$$\left(x',y'\right)=\left(x_1-\frac{2a(ax_1+by_1+c)}{a^2+b^2},\;y_1-\frac{2b(ax_1+by_1+c)}{a^2+b^2}\right)$$

with $a=b=1,c=2$ and $\left(x_1,y_1\right)=\left(\frac12,\frac12\right)$.

Now,

$$ax_1+by_1+c=\frac12+\frac12+2=3$$

So the reflected center is

$$\left(\frac12-\frac{2\cdot1\cdot3}{2},\frac12-\frac{2\cdot1\cdot3}{2}\right)=\left(-\frac52,-\frac52\right)$$

The original radius satisfies

$$r^2=\left(\frac12\right)^2+\left(\frac12\right)^2=\frac12$$

Therefore the image circle is

$$\left(x+\frac52\right)^2+\left(y+\frac52\right)^2=\frac12$$

Expanding,

$$x^2+y^2+5x+5y+12=0$$

Therefore, the correct option is A.