JEE Main 2020 — Circle Question with Solution

Equation of family of circle touching y-axis at

(0, 4) is given by (x – 0)² + (y – 4)² + $$\lambda$$x = 0.

$$\because$$ It passes through (2, 0)

$$\Rightarrow$$ $$\lambda$$ = –10.

$$\Rightarrow$$ Required circle is (x – 0)² + (y – 4)² – 10x = 0

$$\Rightarrow$$ x² + y² – 10x – 8y + 16 = 0

$$\therefore$$ center of circle (5, 4) and radius = 5

By checking all options you can see 4x + 3y – 8 = 0 is not a tangent to the circle.

As distance of 4x + 3y – 8 = 0 from (5, 4)

= $$\left|\frac{24}{5}\right|$$ $$\ne$$ radius