JEE Main 2019 — Mathematics Coordinate Geometry

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

Hence, the radius of the circle is $=h.$=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 ${x}^{2}+{y}^{2}={r}^{2}$x2+y2=r2 is $(0,0)$(0,0) and $1$1 respectively.

Now, since the circle with centre $(h,k)$(h,k) and radius $h$h touches ${x}^{2}+{y}^{2}=1$x2+y2=1externally, then

$$\Rightarrow h+1=\sqrt{{h}^{2}+{k}^{2}}$$⇒h+1=h2+k2​

Squaring both the sides, we get

$$\Rightarrow {h}^{2}+2h+1={h}^{2}+{k}^{2}$$⇒h2+2h+1=h2+k2

$$\Rightarrow {k}^{2}=2h+1$$⇒k2=2h+1

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

$${y}^{2}=2x+1.$$y2=2x+1.

Hence, the required locus is ${y}^{2}=2x+1$y2=2x+1

$$\Rightarrow y=\sqrt{1+2x},x\geq 0.$$⇒y=1+2x​,x≥0.