<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: When a circle touches the x-axis, the radius equals the absolute value of the y-coordinate of the center. Use the y-intercept condition to find the radius.
<p>The circle touches the x-axis at a distance 3 from the origin, so the center is at \((3, r)\) or \((3, -r)\) where r is the radius. The circle equation is \((x-3)^2 + (y-r)^2 = r^2\), which expands to \(x^2 + y^2 - 6x + 9 - 2ry = 0\).</p><p>For the y-intercept, set \(x = 0\): \(y^2 - 2ry + 9 = 0\). The difference between roots is \(2\sqrt{7}\), so \((y_1 - y_2)^2 = 4r^2 - 36 = 28\), giving \(r^2 = 16\), thus \(r = 4\).</p><p>The two circles are \(x^2 + y^2 - 6x - 8y + 9 = 0\) (center \((3, 4)\)) and \(x^2 + y^2 - 6x + 8y + 9 = 0\) (center \((3, -4)\)).</p><p>∴ Answer is (a) and (c).</p>
Correct Answer: A, C