Circles
Tangents from external point
Grade 11
Question:
<p><b>Paragraph for Question nos. 616 and 617</b><br>Consider a circle \(S: x^2 + y^2 - 6x - 4y - 3 = 0\) with centre \(C\) and \(P\) be the point \((-1, -1)\). Also \(PA\) and \(PB\) are tangents drawn to the circle \(S\).</p><p>Radius of the circle circumscribing the triangle \(PAB\) is:</p>
<p>(a) 1</p>
<p>(b) \(\dfrac{3}{2}\)</p>
<p>(c) \(\dfrac{4}{3}\)</p>
<p>(d) \(\dfrac{5}{2}\)</p>
Step-by-Step Solution
Key Concept: The circumradius of triangle PAB equals half the distance PC, since P and C are endpoints of a diameter of the circumcircle (the circle through P, A, B has PC as diameter because ∠PAC = ∠PBC = 90°).
<p><strong>Step 1:</strong> Rewrite circle S in standard form: x² + y² - 6x - 4y - 3 = 0</p><p>(x - 3)² + (y - 2)² = 9 + 4 + 3 = 16</p><p>Centre C = (3, 2), Radius r = 4</p><p><strong>Step 2:</strong> Since PA and PB are tangents from external point P(-1, -1) to circle S:</p><p>∠PAC = ∠PBC = 90° (radius ⊥ tangent)</p><p><strong>Step 3:</strong> Points A, P, B, C form a cyclic quadrilateral where ∠PAC = ∠PBC = 90°</p><p>Both angles subtend the same chord PC, and since both are 90°, PC must be a diameter of the circumcircle of triangle PAB.</p><p><strong>Step 4:</strong> Calculate PC:</p><p>PC = √[(3-(-1))² + (2-(-1))²] = √[16 + 9] = √25 = 5</p><p><strong>Step 5:</strong> Circumradius R = PC/2 = 5/2</p><p>∴ Answer: D</p>
Correct Answer: D