A) True
B) False
C) Can't say
D) Partially True/False
Correct Answer: A
Solution :
| First, we draw a circle and a point from the given information.\ |
| Now, distance between origin i.e. O(0, 0) and \[P\left( 5,\,0 \right),\,OP=\sqrt{{{\left( 5-0 \right)}^{2}}+{{\left( 0-0 \right)}^{2}}}\] |
| [\[\because\]Distance between two points \[\left( {{x}_{1}},\,{{y}_{1}} \right)\] |
| and \[\left( {{x}_{2}},\,{{y}_{2}} \right)\], \[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]] |
| \[=\sqrt{{{5}^{2}}+{{0}^{2}}}=5\]units |
| = Radius of circle and distance between origin O(0, 0) |
| and \[Q\left( 6,\,8 \right),\,OQ=\sqrt{{{\left( 6-0 \right)}^{2}}+{{\left( 8-0 \right)}^{2}}}\] |
| \[=\sqrt{{{6}^{2}}+{{8}^{2}}}=\sqrt{36+64}\] |
| \[=\sqrt{100}=10\]units |
| 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 \[Q\left( 6,\,8 \right)\] , lies outside the circle. |
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