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प्रश्न
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.
पर्याय
True
False
उत्तर
This statement is True.
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((x_2 - x_1)^2 + (y_2 - y_1)^2`
OP = `sqrt((5 - 0)^2 + (0 - 0)^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 > OP
Hence, it is true that point Q(6, 8), lies outside the circle.
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संबंधित प्रश्न
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Find the distance between the points
A(-6,-4) and B(9,-12)
Determine whether the points are collinear.
L(–2, 3), M(1, –3), N(5, 4)
Find the distance between the following pairs of point in the coordinate plane :
(13 , 7) and (4 , -5)
Find the distance of a point (7 , 5) from another point on the x - axis whose abscissa is -5.
Find the distance of a point (12 , 5) from another point on the line x = 0 whose ordinate is 9.
Find the coordinates of O, the centre passing through A( -2, -3), B(-1, 0) and C(7, 6). Also, find its radius.
Calculate the distance between A (5, -3) and B on the y-axis whose ordinate is 9.
Show that the point (0, 9) is equidistant from the points (– 4, 1) and (4, 1)
Find distance between points P(– 5, – 7) and Q(0, 3).
By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(square + square)`
= `sqrt(125)`
= `5sqrt(5)`