JEE Main 2019 — Circle Question with Solution

For this circle x ² + y² = 4

Center C₁ = (0, 0) and radius r₁ = 2

For this circle x² + y² + 6x + 8y – 24 = 0

Center C₂ = (-3, -4) and radius r₂ = $$\sqrt{9+16+24}$$ = 7

So distance between center,

C₁C₂ = 5

r₁ + r₂ = 7

|r₁ - r₂| = 5

As C₁C₂ = |r₁ - r₂|

$$\therefore$$ Circles touches internally.

Here Equation of common tangent is same as common chord.

$$\therefore$$ S₁ - S₂ = 0

$$\Rightarrow$$ (x ² + y² - 4) - (x² + y² + 6x + 8y - 24) = 0

$$\Rightarrow$$ - 6x - 8y + 20 = 0

$$\Rightarrow$$ 3x + 4y - 10 = 0

By checking all the options you can see that,

(6, –2) point satisfy the equation 3x + 4y - 10 = 0.