QJEEM20Ma84f2: JEE Main 2020 Mathematics - Class 11 JEEM20Ma84f2 Solution: If the curves, x 2 – 6x + y 2 + 8 = 0 and x 2 – 8y + y 2 + 16 – …

C₁ : x² + y² – 6x + + 8 = 0

C₁(3, 0) and r₁ = 1

C₂ : x² + y² – 8y + 16 – k = 0

C₂(0, 4) and r₂ = $$\sqrt{k}$$

Two circles touch each other

$$\therefore$$ C₁C₂ = | r₁ $$\pm$$ r₂ |

$$\Rightarrow$$ 5 = | 1 $$\pm$$ $$\sqrt{k}$$ |

$$\therefore$$ 1 + $$\sqrt{k}$$ = 5 or $$\sqrt{k}$$ - 1 = 5

$$\Rightarrow$$ k = 16 or k = 36

So largest value of k = 36.