<p>Let \(\tau\) be a circle with centre \(C(3, 5)\) and \(PA\) and \(PB\) are pair of tangents drawn from an external point \(P(9, 11)\) to the circle \(\tau\). Find the distance between the origin and the point inside the quadrilateral \(ACBP\) which is equidistant from its four vertices.</p>
Step-by-Step Solution
Key Concept: The point equidistant from all four vertices of quadrilateral ACBP is the circumcenter of ACBP. Since PA and PB are tangents from P to circle with center C, angles CAP and CBP are right angles, making ACBP a cyclic quadrilateral with CP as diameter. The circumcenter lies at the midpoint of CP.
<p><strong>Step 1:</strong> Identify the quadrilateral ACBP. Since PA and PB are tangents from external point P to circle τ with center C, we have ∠CAP = ∠CBP = 90°.</p><p><strong>Step 2:</strong> Since ∠CAP = ∠CBP = 90°, both A and B lie on a circle with diameter CP (angles in a semicircle). Therefore ACBP is a cyclic quadrilateral with CP as diameter.</p><p><strong>Step 3:</strong> The point equidistant from all four vertices of a cyclic quadrilateral is its circumcenter, which is the midpoint of the diameter CP.</p><p><strong>Step 4:</strong> Calculate the midpoint M of CP where C(3, 5) and P(9, 11):<br/>M = ((3+9)/2, (5+11)/2) = (6, 8)</p><p><strong>Step 5:</strong> Find the distance from origin O(0, 0) to M(6, 8):<br/>d = √[(6-0)² + (8-0)²] = √(36 + 64) = √100 = 10</p><p>∴ Answer: <strong>10</strong></p>
Correct Answer: 10