Circles
Chord and midpoint locus
Grade 11
Question:
<p><strong>20.</strong> Consider the circle \(x^2 + y^2 = 25\) and a point \(A(1, 2)\) lying inside it. Next consider secants of the circle passing through point \(A\). It turns out that the mid-point of the secants, lie on another circle of centre \((a, b)\) and radius \(r\). Then triplet \((a, b, r)\) is:</p>
<p>(a) \((1, 2, 5)\)</p>
<p>(b) \(\left(\dfrac{1}{2}, 1, \dfrac{\sqrt{5}}{2}\right)\)</p>
<p>(c) \((0, 0, \sqrt{5})\)</p>
<p>(d) \((1, 2, \sqrt{5})\)</p>
Step-by-Step Solution
Key Concept: The locus of midpoints of chords passing through a fixed interior point P forms a circle whose center lies on the line OP (O being the original center), and can be found using the property that OP is perpendicular to the chord at its midpoint.
<p><strong>Step 1:</strong> Let M(h, k) be the midpoint of a secant through A(1, 2) on circle x² + y² = 25.</p><p><strong>Step 2:</strong> For any chord with midpoint M, the line from center O(0,0) to M is perpendicular to the chord. Since the chord passes through A(1, 2) and has midpoint M(h, k), we have: OM ⊥ AM.</p><p><strong>Step 3:</strong> Vector OM = (h, k) and vector AM = (h - 1, k - 2).</p><p><strong>Step 4:</strong> For perpendicularity: OM · AM = 0</p><p>h(h - 1) + k(k - 2) = 0</p><p>h² - h + k² - 2k = 0</p><p><strong>Step 5:</strong> Rearranging: (h - 1/2)² + (k - 1)² = 1/4 + 1 = 5/4</p><p><strong>Step 6:</strong> This is a circle with center (a, b) = (1/2, 1) and radius r = √(5/4) = √5/2.</p><p>∴ Answer: (a, b, r) = (1/2, 1, √5/2)</p>
Correct Answer: B