JEE Main 2016 — Circle Question with Solution

EF = perpendicular bisector of chord AB

BG = perpendicular to y-axis

Here C = center of the circle

mid-point of chord AB, D = ($$-$$ 1, 3)

slope of AB = $$\frac{4-2}{-2-0}$$ = $$-$$ 1

$$\because$$   EF  $$\perp$$ AB

$$\therefore$$   Slope of EF = 1

Equation of EF, y $$-$$ 3 = 1 (x + 1)

$$\Rightarrow$$    y = x + 4       . . . .(i)

Equation of BG

y = 2         . . . . (ii)

From equations (i) and (ii)

x = $$-$$ 2, y = 2

since C be the point of intersection of EF and BG, therefore center, C = ($$-$$ 2, 2)

Now coordinates of center C satiesfy the equation

2x $$-$$ 3y + 10 = 0

Hence 2x $$-$$ 3y + 10 = 0   is the equation of the diameter