<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 coordinates 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: Find the center of the circle.</strong></p><p>A circle making intercepts of length 2a on the x-axis and 2b on the y-axis, and passing through the origin, has its diameter endpoints at (2a, 0) and (0, 2b) on the respective axes.</p><p>The center of the circle is the midpoint of the chord joining (2a, 0) and (0, 2b): Center = (a, b)</p><p><strong>Step 2: Find the radius vector at origin.</strong></p><p>The radius at point (0, 0) is from center (a, b) to (0, 0).</p><p>Direction vector of radius = (0 - a, 0 - b) = (-a, -b)</p><p>Slope of radius = b/a</p><p><strong>Step 3: Find the equation of tangent.</strong></p><p>The tangent at any point on a circle is perpendicular to the radius at that point.</p><p>If the radius has slope b/a, the tangent has slope -a/b.</p><p>Tangent passes through (0, 0) with slope -a/b:</p><p>y - 0 = (-a/b)(x - 0)</p><p>by = -ax</p><p>ax + by = 0</p><p><strong>Step 4: Verify.</strong></p><p>The equation ax + by = 0 represents a line through the origin perpendicular to the radius from (a, b) to (0, 0), which confirms it is the correct tangent.</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a