Question:
<p>The circle passing through (1, - 2) and touching the axis of x at (3, 0) also passes through the point</p>
<p style="display:inline">(2, -5)</p>
<p style="display:inline">(-2, 5)</p>
<p style="display:inline">(5, -2)</p>
<p style="display:inline">(-5, 2)</p>
Step-by-Step Solution
Key Concept: When a circle touches the x-axis at point (3, 0), its center must lie on the vertical line x = 3, so the center is at (3, k) for some k. The radius equals |k|. Use the condition that the circle passes through (1, -2) to find k: (1-3)² + (-2-k)² = k².
<p>Let the equation of circle be (x - 3)<sup>2</sup> + (y - 0)<sup>2</sup> + <span class="math-tex">$\lambda y$</span> = 0<br />
<img alt="" data-imgur-src="x3QQttC.png" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/x3QQttC.png" style="width: 150px; height: 103px;" /><br />
As it passes through (1, -2)<br />
<span class="math-tex">$\therefore$</span> (1 - 3)<sup>2</sup> + (-2)<sup>2 </sup>+ <span class="math-tex">$\lambda$</span> (-2) = 0<br />
<span class="math-tex">$\Rightarrow$</span> 4 + 4 - 2<span class="math-tex">$\lambda$</span> = 0 <span class="math-tex">$\Rightarrow$</span><span class="math-tex">$\lambda$</span> = 4<br />
<span class="math-tex">$\therefore$</span> Equation of circle is (x - 3)<sup>2</sup> + y<sup>2</sup> + 4y = 0<br />
By hit and trial method, we see that point (5, - 2) satisfies equation of circle.</p>
Correct Answer: C