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A circle has its Centre at the origin and a point P (5, 0) lies on it. The point Q (6, 8) lies outside the circle.

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Answer.     [True]
Solution.    The centre of the circle is O (0, 0).
If point P(5, 0) lies on the circle then the distance between O(0, 0) and P(5, 0) is the radius of the circle
OP= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}
OP= \sqrt{\left ( 5-0 \right )^{2}+\left ( 0-0 \right )^{2}}
OP= \sqrt{\left ( 5 \right )^{2}}= 5
OP = 5
Radius of circle = 5
If point Q (6, 8) is outside the circle then the distance between O(0, 0) and Q(6, 8) is grater then the radius of the circle
(x1, y1) = (0, 0)                        (x2, y2) = (6, 8)
OQ= \sqrt{\left ( 6-0 \right )^{2}+\left ( 8-0 \right )^{2}}
OQ= \sqrt{36+64}
         = \sqrt{100}=10

Here point OQ is greater than the radius of circle
Hence, point Q(6, 8) lies outside the circle

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