JEE Mathematics 2026 Question with Solution

Given: $P$ lies on the circle $x^2 + y^2 = 4$, $Q$ lies on the line $5x + y + 2 = 0$, and $x - y + 1 = 0$ is the perpendicular bisector of $PQ$.

Find: $13$ times the sum of abscissas of all such points $P$.

Since $x - y + 1 = 0$ is the perpendicular bisector of $PQ$, point $Q$ is the reflection of point $P$ in this line.

Let $P = (x_1, y_1)$. For the line $x - y + 1 = 0$, the image formula gives

$$\frac{x_2 - x_1}{1} = \frac{y_2 - y_1}{-1} = -2\frac{x_1 - y_1 + 1}{1^2 + (-1)^2} = -(x_1 - y_1 + 1)$$

So,

$$x_2 = x_1 - (x_1 - y_1 + 1) = y_1 - 1$$

and

$$y_2 = y_1 + (x_1 - y_1 + 1) = x_1 + 1$$

Hence $Q = (y_1 - 1, x_1 + 1)$.

Since $Q$ lies on $5x + y + 2 = 0$,

$$5(y_1 - 1) + (x_1 + 1) + 2 = 0$$ $$x_1 + 5y_1 - 2 = 0$$ $$x_1 + 5y_1 = 2$$

Also, $P$ lies on the circle, so

$$x_1^2 + y_1^2 = 4$$

Using

$$y_1 = \frac{2 - x_1}{5}$$

in the circle equation,

$$x_1^2 + \left(\frac{2 - x_1}{5}\right)^2 = 4$$ $$25x_1^2 + 4 - 4x_1 + x_1^2 = 100$$ $$26x_1^2 - 4x_1 - 96 = 0$$

Dividing by $2$,

$$13x_1^2 - 2x_1 - 48 = 0$$

If the two possible abscissas are the roots of this quadratic, then their sum is

$$-\frac{b}{a} = \frac{2}{13}$$

Therefore,

$$13 \times \frac{2}{13} = 2$$

So the required value is $2$.

The solution also shows a page-level correct answer of $20$, but the worked solution leads to $2$. Hence the answer is determined from the solution working.

Given: Reflection is taken about the line $x - y + 1 = 0$.

Find: A quicker way to form the condition on $P$.

Rewrite the line as

$$y = x + 1$$

For reflection about a line of the form $y = x + c$, the image of $(x, y)$ is $(y-c, x+c)$. So the image of $P(x_1, y_1)$ is directly

$$Q = (y_1 - 1, x_1 + 1)$$

Now use the condition that $Q$ lies on

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

Substituting gives

$$5(y_1 - 1) + (x_1 + 1) + 2 = 0$$ $$x_1 + 5y_1 = 2$$

Together with

$$x_1^2 + y_1^2 = 4$$

this leads to the quadratic

$$13x_1^2 - 2x_1 - 48 = 0$$

The sum of the abscissas is therefore

$$\frac{2}{13}$$

and the required value is

$$13 \times \frac{2}{13} = 2$$

Therefore, the final answer is $2$.