A circle has centre at the origin and a point \(P(5,0)\) lies on it. The point \(Q(6,8)\) lies outside the circle.
True.
Step 1: The circle is centered at the origin \(O(0,0)\). Since the point \(P(5,0)\) lies on the circle, the distance from the center to \(P\) is the radius.
Step 2: Calculate the radius:
\(OP = \sqrt{(5-0)^2 + (0-0)^2} = \sqrt{25} = 5\,\text{units}\).
Step 3: Now calculate the distance from the center \(O(0,0)\) to the point \(Q(6,8)\):
\(OQ = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,\text{units}\).
Step 4: Compare the distances:
The radius = 5 units, but the distance \(OQ = 10\,\text{units}\).
Step 5: Since \(10 > 5\), point \(Q\) lies outside the circle.