JEE 2023 — Mathematics PYQ

Circle ka centre $(2,0)$ hai, toh uski equation hogi:

Yahan se $y^2=r^2-(x-2{)}^2$.

Step 2: Ellipse ki equation se $y^2$ nikalna

Ellipse hai: $\frac{x^2}{36}+\frac{y^2}{16}=1$

$y^2=16\left(1-\frac{x^2}{36}\right)=16-\frac{4x^2}{9}$

Step 3: Inscribed condition (Dono ko solve karein)

Circle ellipse ke andar hai, toh intersection point ke liye $y^2$ ki values barabar rakhein:

Kyunki circle "largest" hai aur inscribed hai, toh ye quadratic equation $x$ ke liye ek perfect square honi chahiye (yaani $D=0$).

$D=b^2-4ac=0$

Step 4: Point $(1,\alpha )$ circle par hai

Circle ki equation mein $(1,\alpha )$ rakhein:

Step 5: Final value nikalna

Hume $10{\alpha }^2$ nikalna hai:

Final Answer:

Iska sahi jawab 118 hai.