<p>Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length \(2\sqrt{7}\) on y-axis is(are):</p>
<p>(a) \(x^2 + y^2 - 6x + 8y + 9 = 0\)</p>
<p>(b) \(x^2 + y^2 - 6x + 7y + 9 = 0\)</p>
<p>(c) \(x^2 + y^2 - 6x - 8y + 9 = 0\)</p>
<p>(d) \(x^2 + y^2 - 6x - 7y + 9 = 0\)</p>
Step-by-Step Solution
Key Concept: Use the tangency condition (circle touches x-axis) to constrain the center, then use the y-intercept condition to find the exact center location.
<p>Circle touches x-axis at distance 3 from origin, so it touches at \((3, 0)\) or \((-3, 0)\). The center is at \((3, h)\) or \((-3, h)\) with radius \(|h|\). For intercept on y-axis of length \(2\sqrt{7}\): if circle equation is \((x-3)^2 + (y-h)^2 = h^2\), at \(x=0\): \(9 + (y-h)^2 = h^2\), giving \(y^2 - 2hy + 9 = 0\). The intercept is \(|y_1 - y_2| = 2\sqrt{h^2 - 9} = 2\sqrt{7}\), so \(h^2 = 16\), thus \(h = ±4\). Centers are \((3, 4)\), \((3, -4)\). The equations are \(x^2 + y^2 - 6x + 8y + 9 = 0\) and \(x^2 + y^2 - 6x - 8y + 9 = 0\).</p><p>∴ Answer is (a, c).</p>
Correct Answer: a, c