Question

Question 7
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

True
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), $PO=\sqrt{(5-0{)}^2+(0-0^2)}$
$\left[\begin{array}{c}\because \text{distance between two points}(x_1,y_1)and(x_2,y_2)d\\ =\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\end{array}\right]=\sqrt{5^2+0^2}=5=Radiusofcircle.\text{And Q(6,8)},$

$\text{OQ}=\sqrt{(6-0{)}^2+(8-0{)}^2}=\sqrt{(6^2+8^2)}=\sqrt{36+64}=\sqrt{100}=10$

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>OP
Hence, it is true that point Q(6,8), lies outside the circle.