Question:
<p>The equation of the circle which touches x-axis at (3, 0) and passes through (1, 4) is given by</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - 6x - 5y + 9 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> + 6x + 5y - 9 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> + 6x - 5y + 9 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - 6x + 5y - 9 = 0</p>
Step-by-Step Solution
Key Concept: A circle touching the x-axis at (h, 0) has its center at (h, k) and its radius equal to the absolute value of k.
<html><body><p><img alt="" data-imgur-src="h7rVq6s.png" height="119" src="https://media-mycbseguide.s3.amazonaws.com/images/imgur/1623761511-48ev5j.jpg" width="149"/><br/>
Since, the circle touches x-axis at (3, 0).<br/>
<span class="math-tex">$\Rightarrow$</span> centre of the circle is (3, k).<br/>
Now, CA<sup>2</sup> = CB<sup>2</sup><br/>
<span class="math-tex">$\Rightarrow$</span> (3 - 3)<sup>2</sup> + (k - 0)<sup>2</sup><br/>
= (3 - 1)<sup>2</sup> + (k - 4)<sup>2</sup><br/>
<span class="math-tex">$\Rightarrow$</span> k<sup>2</sup> = 4 + k<sup>2</sup> - 8k + 16<br/>
<span class="math-tex">$\Rightarrow k=\frac{5}{2}$</span><br/>
Required equation is<br/>
<span class="math-tex">$(x-3)^{2}+\left(y-\frac{5}{2}\right)^{2}=\left(\frac{5}{2}\right)^{2}$</span><br/>
<span class="math-tex">$\Rightarrow$</span> x<sup>2</sup> + y<sup>2</sup> - 6x - 5y + 9 = 0</p></body></html>
Correct Answer: A