JEE Mathematics 2025 Question with Solution

Given: The circle is $x^2 + (y - 1)^2 = 2$ and the line $x + y = 3$ is tangent to the circle at $P(x_1,y_1)$. The same line touches the ellipses at $Q(x_2,y_2)$ and $R(x_3,y_3)$. Also, $P$ is the midpoint of $QR$ and $PQ = \frac{2\sqrt{2}}{3}$.

Find: The value of $9(x_1y_1 + x_2y_2 + x_3y_3)$.

The tangency point on the circle is the foot of the perpendicular from the center $(0,1)$ to the line $x+y-3=0$.

So,

Now the line $x+y=3$ has direction vector $(1,-1)$, so a unit direction vector is

Since $P$ is the midpoint of $QR$ and $PQ = \frac{2\sqrt{2}}{3}$,

Then

Hence,

and

Therefore,

Now add:

Thus,

Therefore, the required value is $46$, so the correct option is A.

Given: The common tangent is $x+y=3$, the point of contact with the circle is $P$, and $P$ is the midpoint of $QR$ with $PQ=\frac{2\sqrt{2}}{3}$.

Find: $9(x_1y_1+x_2y_2+x_3y_3)$.

First find $P$. The circle has center $(0,1)$ and radius $\sqrt{2}$. Since the line is tangent,

so the line is indeed tangent. The radius to the tangency point is perpendicular to the line, hence $P$ is the projection of $(0,1)$ on $x+y=3$, which gives

Thus,

Now move along the tangent line. A unit vector parallel to $x+y=3$ is

Because $P$ is the midpoint of $QR$ and the distance from $P$ to each endpoint is $PQ=\frac{2\sqrt{2}}{3}$, the shift from $P$ to either point is

So,

Now compute the products:

Therefore,

Hence,

Therefore, the correct option is A.