There is a circle with the center [tex](6,-8)[/tex] and radius = [tex]20[/tex] units
We have to find whether the points given in the options will not lie inside the circle if the radius of the circle is halved.
"If a point lies inside the circle, distance between the center and the given point will be less than the radius of the circle".
To find the distance between two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is given by,
Distance = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
If a point lies inside the circle with radius 10 units, distance between the point and the center will be less than 10 units.
Option A
Distance between the points [tex](-\frac{1}{20},0)[/tex] and [tex](6, -8)[/tex]
= [tex]\sqrt{(6+\frac{1}{20})^2+(-8-0)^2}[/tex]
= [tex]\sqrt{100.6025}[/tex]
= [tex]10.03[/tex] units
Since, [tex]10.03>10[/tex] units
Therefore, [tex](-\frac{1}{20},0)[/tex] will lie outside the circle.
Option B
Distance between [tex](\frac{1}{20},0)[/tex] and [tex](6,-8)[/tex] = [tex]\sqrt{(6-\frac{1}{20})^2+(-8-0)^2}[/tex]
= [tex]\sqrt{99.4025}[/tex]
= [tex]9.97[/tex] units
Since, [tex]9.97<10[/tex] units
Therefore, [tex](\frac{1}{20},0)[/tex] will lie inside the circle.
Option C
Distance between [tex](-\frac{1}{10},-6)[/tex] and [tex](6,-8)[/tex] = [tex]\sqrt{(6+\frac{1}{10})^2+(-6+8)^2}[/tex]
= [tex]\sqrt{41.21}[/tex]
= [tex]6.42[/tex] units
Since, [tex]6.42<10[/tex] units
Therefore, [tex](-\frac{1}{10},-6)[/tex] will lie inside the circle.
Option D
Distance between [tex](8,10)[/tex] and [tex](6,-8)[/tex] = [tex]\sqrt{(8-6)^2+(10+8)^2}[/tex]
= [tex]\sqrt{328}[/tex]
= [tex]18.11[/tex] units
Since, [tex]18.11>10[/tex] units
Therefore, [tex](8, 10)[/tex] will lie outside the circle.
Now we can conclude that [tex](-\frac{1}{20},0)[/tex] and [tex](8,10)[/tex] will lie outside the circle.
Options (A) and (D) will be the correct options.
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