<p>A circle with centre C(3, 5) has AB as diameter, and P(9, 11) is a point outside the circle such that PA and PB are tangents to the circle. M(6, 8) is the midpoint of PC. The point inside the quadrilateral ACBP which is equidistant from all four vertices is M(6, 8). Find the distance from the origin to the point M.</p>
Step-by-Step Solution
Key Concept: If M is equidistant from all four vertices of quadrilateral ACBP, then M is the circumcenter. Since AB is a diameter with center C, and PA, PB are tangents, ACBP forms a cyclic quadrilateral where M is the circumcenter. Use the distance formula from origin to M.
<p><strong>Step 1:</strong> Recognize that since M is equidistant from all four vertices A, C, B, P, point M is the circumcenter of quadrilateral ACBP.</p><p><strong>Step 2:</strong> Given M(6, 8) and we need to verify the configuration: C(3, 5) is the center of the circle with diameter AB. Since PA and PB are tangents from external point P(9, 11), angles CAP and CBP are right angles (radius ⊥ tangent).</p><p><strong>Step 3:</strong> In cyclic quadrilateral ACBP where ∠CAP = ∠CBP = 90°, the circumcenter M must be equidistant from all vertices. This is consistent with M being the midpoint of the diagonal CP (since ∠CAP = ∠CBP = 90° means A and B lie on a circle with CP as diameter).</p><p><strong>Step 4:</strong> Calculate distance from origin O(0, 0) to M(6, 8):<br/>Distance = √[(6-0)² + (8-0)²]<br/>Distance = √[36 + 64]<br/>Distance = √100<br/>Distance = 10</p><p>∴ Answer: <strong>10</strong></p>
Correct Answer: 10