JEE Main 2022 — Circle Question with Solution

$$L_1:4x+3y+2=0$$

$$L_2:3x-4y-11=0$$

Since circle C touches the line L₂ at Q intersection point Q of L₁ and L₂, is (1, $$-$$2)

$$\because$$ P lies of L₁

$$\therefore$$ $$P\left(x,-\frac{1}{3}(2+4x)\right)$$

Now, $$PQ=5\Rightarrow (x-1{)}^2+{\left(\frac{4x+2}{3}-2\right)}^2=25$$

$$\Rightarrow (x-1{)}^2\left[1+\frac{16}{9}\right]=25$$

$$\Rightarrow (x-1{)}^2=9$$

$$\Rightarrow x=4,-2$$

$$\because$$ Circle lies below the x-axis

$$\therefore$$ y = $$-$$6

P(4, $$-$$6)

Now distance of P from 5x $$-$$ 12y + 51 = 0

$$=\left|\frac{20+72+51}{13}\right|=\frac{143}{13}=11$$