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.

Answer

1

Explanation

First , we draw a circle and a point from the given information.

Now , distance between origin i.e., O(0,0) and P (5,0), OP `=sqrt((5-0)^(2)+(0-0)^(2))`
`[ :'` Distance between two points `(x_(1),y_(1))and (x_(2),y_(2)),d= sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))]`
`=sqrt(5^(2)+0^(2))=5=` Radius of circle and distance between origin O(0,0)
and Q(6,8),OQ`=sqrt((6-0)^(2)+(8-0)^(2))=sqrt(6^(2)+8^(2))=sqrt(36+64)=sqrt(100)=10`
We know that , if the distance of any point from the centre is less than/ equal to/more than the radius , then the point is inside/on/outside the circle, respectively.
Here , we see that , `OQ lt OP`
Hence , it is true that point Q (6,8) , lies outside the circle.