Question:
<p>The equation of the circle having centre (1, -2) and passing through the point of intersection of lines 3x + y = 14, 2x + 5y = 18 is</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - 2x - 4y - 20 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> + 2x - 4y - 20 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> - 2x + 4y - 20 = 0</p>
<p style="display:inline">x<sup>2</sup> + y<sup>2</sup> + 2x + 4y - 20 = 0</p>
Step-by-Step Solution
Key Concept: Find the intersection point of the lines to identify a point on the circle, then use its distance from the given center to determine the radius for the standard circle equation.
<p>The point of intersection of 3x + y - 14 = 0 and 2x + 5y - 18 = 0 is (4, 2).<br />
Centre of the circle is (1, -2).<br />
<span class="math-tex">\(\Rightarrow\)</span> radius = <span class="math-tex">\(\sqrt{(4-1)^{2}+(2+2)^{2}}=5\)</span><br />
The equation of the circle is<br />
(x - 1)<sup>2</sup> + (y + 2)<sup>2</sup> = 5<sup>2</sup><br />
<span class="math-tex">\(\Rightarrow\)</span> x<sup>2</sup> + y<sup>2</sup> - 2x + 4y - 20 = 0</p>
Correct Answer: C