Question:
<p>The locus of the centres of the circles, which touch the circle, x<sup>2</sup> + y<sup>2</sup> = 1 externally, also touch the Y-axis and lie in the first quadrant, is</p>
<p style="display:inline"><span class="math-tex">\(x=\sqrt{1+2 y}, y \geq 0\)</span></p>
<p style="display:inline"><span class="math-tex">\(y=\sqrt{1+4 x}, x \geq 0\)</span></p>
<p style="display:inline"><span class="math-tex">\(x=\sqrt{1+4 y}, y \geq 0\)</span></p>
<p style="display:inline"><span class="math-tex">\(y=\sqrt{1+2 x}, x \geq 0\)</span></p>
Step-by-Step Solution
Key Concept: The locus is determined by setting the distance between the center (h, k) and the origin equal to the sum of the circle's radius h and the fixed circle's radius 1.
<p>Let (h, k) be the centre of the circle and radius r = h, as circle touch the Y-axis and other circle x<sup>2</sup> + y<sup>2</sup> = 1 whose centre (0, 0) and radius is 1<br />
<img alt="" data-imgur-src="inm1k5f.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/inm1k5f.png" style="width: 200px; height: 158px;" /><br />
<span class="math-tex">$\therefore$</span> OC = r + 1 [<span class="math-tex">$\because$</span> if circles touch each other externally, then C<sub>1</sub>C<sub>2</sub> = r<sub>1</sub> + r<sub>2</sub>]<br />
<span class="math-tex">$\Rightarrow \sqrt{h^{2}+k^{2}}$</span> = h + 1, h > 0 and k > 0, for first quadrant<br />
<span class="math-tex">$\Rightarrow$</span> h<sup>2</sup> + k<sup>2</sup> = h<sup>2</sup> + 2h + 1<br />
<span class="math-tex">$\Rightarrow$</span> k<sup>2</sup> = 2h + 1<br />
<span class="math-tex">$\Rightarrow$</span> k = <span class="math-tex">$\sqrt{1+2 h}$</span>, as k > 0<br />
Now, on taking locus of centre (h, k), we get<br />
y <span class="math-tex">$=\sqrt{1+2 x}, x \geq 0$</span></p>
Correct Answer: D