Question:
<p>From a variable point P, tangents are drawn one each to the two circles x<sup>2</sup> + y<sup>2</sup> = a<sup>2</sup>, x<sup>2</sup> + y<sup>2</sup> = b<sup>2</sup>, a > b. If the tangents are mutually perpendicular, then P describes a</p>
<p style="display:inline">circle with radius a<sup>2</sup> + b<sup>2</sup></p>
<p style="display:inline">circle with radius <span class="math-tex">\(\sqrt{a^{2}+b^{2}}\)</span></p>
<p style="display:inline">circle with radius <span class="math-tex">\(\sqrt{\frac{a^{2}+b^{2}}{2}}\)</span></p>
<p style="display:inline">circle with radius <span class="math-tex">\(\frac{1}{2} \sqrt{a^{2}+b^{2}}\)</span></p>
Step-by-Step Solution
Key Concept: The locus is found by applying the tangency condition to two lines with perpendicular slopes and eliminating the slope parameter by summing the resulting equations.
<html><body><p><img alt="" data-imgur-src="QDHSmoS.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1623761675-k6fefg.jpg" style="width: 184px; height: 178px;"/><br/>
Let P = (h, k)<br/>
Let the perpendicular tangents be PB and PA with slopes m and <span class="math-tex">$-\frac{1}{m}$</span> respectively.<br/>
The equation of PB is y - k = m(x - h)<br/>
The perpendicular distance of PB from origin O = |OB|<br/>
<span class="math-tex">$\Leftrightarrow b=\frac{|m h-k|}{\sqrt{1+m^{2}}}$</span> <span class="math-tex">$\Rightarrow$</span> (mh - k)<sup>2</sup> = b<sup>2</sup>(1 + m<sup>2</sup>) ...(i)<br/>
Similarly, OA <span class="math-tex">$\perp$</span> PA gives <span class="math-tex">$a=\frac{|m k+h|}{\sqrt{1+m^{2}}}$</span><br/>
<span class="math-tex">$\Rightarrow$</span> (mk + h)<sup>2</sup> = (1 + m<sup>2</sup>)a<sup>2</sup> ...(ii)<br/>
Adding (i) and (ii), we get<br/>
(1 + m<sup>2</sup>) (h<sup>2</sup> + k<sup>2</sup>) = (1 + m<sup>2</sup>) (a<sup>2</sup> + b<sup>2</sup>)<br/>
<span class="math-tex">$\Rightarrow$</span> P(h, k) satisfies x<sup>2</sup> + y<sup>2</sup> = a<sup>2</sup> + b<sup>2</sup> which represents a circle with radius <span class="math-tex">$\sqrt{a^{2}+b^{2}}$</span></p></body></html>
Correct Answer: B