Question:
<p>If the circles x<sup>2</sup> + y<sup>2 </sup>- 16x - 20y + 164 = r<sup>2 </sup>and <span class="math-tex">\((x-4) ^2+(y-7)^{2}=36\)</span> intersect at two distinct points, then</p>
<p style="display:inline">1 < r < 11</p>
<p style="display:inline">0 < r < 1</p>
<p style="display:inline">r = 11</p>
<p style="display:inline">r > 11</p>
Step-by-Step Solution
Key Concept: Two circles intersect at two distinct points if and only if the distance between their centers is less than the sum of their radii and greater than the absolute difference of their radii.
<p>Circle I is x<sup>2</sup> + y<sup>2</sup> - 16x - 20y + 164 = r<sup>2</sup><br />
<span class="math-tex">$\Rightarrow$</span> (x - 8)<sup>2</sup> + (y - 10)<sup>2</sup> = r<sup>2</sup><br />
<span class="math-tex">$\Rightarrow$</span> C1 (8, 10) is the centre of lst circle and r<sub>1</sub> = r is its radius<br />
Circle II is (x - 4)<sup>2</sup> + (y - 7)<sup>2</sup> = 36<br />
<span class="math-tex">$\Rightarrow$</span> C<sub>2</sub>(4, 7) is the centre of 2nd circle 'and r<sub>2</sub> = 6 is its radius.<br />
Two circles intersects if |r<sub>1</sub> - r<sub>2</sub>| < C<sub>1</sub>C<sub>2</sub> < r<sub>1</sub> + r<sub>2</sub><br />
<span class="math-tex">$\Rightarrow$</span> <span class="math-tex">$|r-6|<\sqrt{(8-4)^{2}+(10-7)^{2}}<r+6$</span><br />
<span class="math-tex">$\Rightarrow$</span> <span class="math-tex">$|r-6|<\sqrt{16+9}<r+6$</span><br />
<span class="math-tex">$\Rightarrow$</span> <span class="math-tex">$|r-6|<5<r+6$</span><br />
Now as, 5 < r+6 always, we have to solve only <br />
<span class="math-tex">$|r-6|<5$</span> <span class="math-tex">$\Rightarrow$</span> <span class="math-tex">$-5<r-6<5$</span><br />
<span class="math-tex">$\Rightarrow$</span> 6 - 5 < r < 5 + 6 <span class="math-tex">$\Rightarrow$</span> 1 < r < 11</p>
Correct Answer: A