The locus of the centres of the circles, which touch the circle, x...

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 $= h .</p><p>Also, we know that if two circles touches externally, then the distance between their centres is equal to the sum of their radii.</p><p>And, the centre and radius of a circle${x}^{2} + {y}^{2} = {r}^{2}$is \left(0, 0\right)$ and $1 respectively.</p><p>Now, since the circle with centre \left(h, k\right) and radius$h$touches${x}^{2} + {y}^{2} = 1$externally, then</p><p>$⇒ h + 1 = \sqrt{{h}^{2} + {k}^{2}}$</p><p>Squaring both the sides, we get</p><p>$⇒ {h}^{2} + 2h + 1 = {h}^{2} + {k}^{2}$</p><p>$⇒ {k}^{2} = 2h + 1

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

${y}^{2} = 2x + 1 .</p><p>Hence, the required locus is${y}^{2} = 2x + 1$</p><p>$⇒ y = \sqrt{1 + 2x}, x \ge 0 .