<p>The equation(s) of the tangent at the point (0, 0) to the circle, making intercepts of lengths 2<i>a</i> and 2<i>b</i> units on the coordinate axes, is/are:</p>
<p>(a) \(ax + by = 0\)</p>
<p>(b) \(ax - by = 0\)</p>
<p>(c) \(x = y\)</p>
<p>(d) \(bx + ay = 0\)</p>
Step-by-Step Solution
Key Concept: A circle passing through the origin with intercepts 2a and 2b on the axes has its center at (a, b), and the tangent at the origin is perpendicular to the radius at that point. The radius from center (a, b) to origin (0, 0) has slope b/a, so the tangent has slope -a/b.
<p><strong>Step 1:</strong> Find the center of the circle. A circle making intercepts of lengths 2a and 2b on the coordinate axes, passing through the origin, has intercepts on the x-axis from (0,0) to (2a,0) and on the y-axis from (0,0) to (0,2b). The center is at the midpoint of the diameter, which is C(a, b).</p><p><strong>Step 2:</strong> Find the slope of the radius at origin. The radius from center C(a, b) to the origin O(0, 0) has slope m_radius = (b - 0)/(a - 0) = b/a.</p><p><strong>Step 3:</strong> Find the slope of the tangent. Since the tangent at any point on a circle is perpendicular to the radius at that point: m_tangent × m_radius = -1, so m_tangent = -a/b.</p><p><strong>Step 4:</strong> Write the tangent equation. The tangent passes through (0, 0) with slope -a/b: y - 0 = (-a/b)(x - 0), which gives y = (-a/b)x, or bx + ay = 0.</p><p><strong>Step 5:</strong> Verify the answer. The equation bx + ay = 0 passes through (0, 0) ✓ and has slope -a/b ✓, confirming it is perpendicular to the radius.</p><p><strong>∴ Answer:</strong> d</p>
Correct Answer: d