JEE Main 2019 — Circle Question with Solution

x² + y² + 4x $$-$$ 6y $$-$$ 12 = 0

Equation of tangent at (1, $$-$$ 1)

x $$-$$ y + 2(x + 1) $$-$$ 3(y $$-$$ 1) $$-$$ 12 = 0

3x $$-$$ 4y $$-$$ 7 = 0

$$\therefore$$   Equation of circle is

(x² + y² + 4x $$-$$ 6y $$-$$ 12) + $$\lambda$$ (3x $$-$$ 4y $$-$$ 7) = 0

It passes through (4, 0) :

(16 + 16 $$-$$ 12) + $$\lambda$$ (12 $$-$$ 7) = 0

$$\Rightarrow$$  20 + $$\lambda$$(5) = 0

$$\Rightarrow$$  $$\lambda$$ = $$-$$ 4

$$\therefore$$  (x² + y² + 4x $$-$$ 6y $$-$$ 12) $$-$$ 4(3x $$-$$ 4y $$-$$ 7) = 0

or   x² + y² $$-$$ 8x + 10y + 16 = 0

Radius = $$\sqrt{16+25-16}=5$$