Let $C$ be the circle $x^{2}+(y-1)^{2}=2$, $E_{1}$ and $E_{2}$ be two ellipses whose centres lie at the origin and major axes lie on x-axis and y-axis respectively. Let the straight line $x+y=3$ touch the curves $C$, $E_{1}$ and $E_{2}$ at $P(x_{1}, y_{1})$, $Q(x_{2}, y_{2})$ and $R(x_{3}, y_{3})$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ=\frac{2\sqrt{2}}{3}$, the value of $9(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})$ is equal to ______.

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Circle

Let

C

be the circle

x^{2} + (y - 1)^{2} = 2

E_{1}

and

E_{2}

be two ellipses whose centres lie at the origin and major axes lie on x-axis and y-axis respectively. Let the straight line

x + y = 3

touch the curves

C

E_{1}

and

E_{2}

P (x_{1}, y_{1})

Q (x_{2}, y_{2})

and

R (x_{3}, y_{3})

respectively. Given that

P

is the mid-point of the line segment

QR

and

P Q = \frac{2 2}{3}

, the value of

9 (x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3})

is equal to ______.

Answer: 46

The line

x + y = 3

touches the circle

x^{2} + (y - 1)^{2} = 2

because the centre

(0, 1)

is at distance

\frac{∣0 + 1 - 3∣}{2} = 2

from it, and the foot of the perpendicular is

P (1, 2)

. Along the line the unit direction is

\frac{1}{2} (1, - 1)

, so the two points at distance

\frac{2 2}{3}

from

P

are

Q (\frac{5}{3}, \frac{4}{3})

and

R (\frac{1}{3}, \frac{8}{3})

. Then

x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} = 2 + \frac{20}{9} + \frac{8}{9} = \frac{46}{9}

, and multiplying by

9

gives

46

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Integer answer