IIT JEE 1983 Maths - 'Adapted' - FIB to Single-Digit Integer Q7

The circle equation can be simplified to $(x - 1)^{2} + (y + 2)^{2} = 5^{2}$ , which describes a circle with center $(1,-2)$ and radius $5$ . Now as $(1,-2)$ lies on the line $4x - 3y - 10 = 0$ , this line will intersect the circle at opposite ends of a diameter, so $(1,-2)$ will be the midpoint of $(a_{1}, a_{2})$ and $(a_{3}, a_{4})$ . Thus

$\left(\dfrac{a_{1} + a_{3}}{2}, \dfrac{a_{2} + a_{4}}{2}\right) = (1, -2) \Longrightarrow a_{1} + a_{3} = 2, a_{2} + a_{4} = -4 \Longrightarrow a_{1} + a_{2} + a_{3} + a_{4} = -2$ .

The given expression then evaluates to $5 + \lfloor \dfrac{-2}{5} \rfloor = 5 + (-1) = \boxed{4}$ .

I am better than kushagra juneja, Understand u, I am in class 10 and solved this problem in 2 min, so just give me cheers We get x=3y+10/4(hahah) Then a eq in y =y^2+4y-12=0 So answer =5+-0.4=5-1=4 Understand all u idiots

$Y=4X/3 -10/3. ~ substituting~ in~ (X- 1)^2+(X+2)^2=25,\\ 9*(X- 1)^2+16(X-1)^2=25*9.\\ \therefore~ (X-1)^2=9,...and..X= -2,4. \\ Y= -6, 2..\\ (a_1, . . .a_4)=(-2, -6, 4, 2).\\ So~5+\lfloor \dfrac{(-2)+(-6)+4+2 } 5 \rfloor=5-1=\huge 4.$