JEE Main 2019 — Circle Question with Solution

Let, the centre of the circle whose locus is to be determined is $P(h, k)$ and it touches $y$-axis in the first quadrant and we know that if a circle touches the $y$-axis, then its radius is equal to the absolute value of the $x$-coordinate of the centre.

Hence, the radius of the circle is $=h.$

Also, we know that if two circles touches externally, then the distance between their centres is equal to the sum of their radii.

And, the centre and radius of a circle $x^2+y^2=r^2$ is $\left(0, 0\right)$ and $1$ respectively.

Now, since the circle with centre $\left(h, k\right)$ and radius $h$ touches $x^2+y^2=1$externally, then

$\Rightarrow h+1=\sqrt{h^2+k^2}$

Squaring both the sides, we get

$\Rightarrow h^2+2h+1=h^2+k^2$

$\Rightarrow k^2=2h+1$

Now, to get the locus, replace $\left(h, k\right)$ by $\left(x, y\right),$ to get

$y^2=2x+1.$

Hence, the required locus is $y^2=2x+1$

$\Rightarrow y=\sqrt{1+2x}, x\ge 0.$