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The center of a circle is (10, –3). The point (10, 9) is out
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06 Oct 2018, 15:08
06 Oct 2018, 15:08
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Question Stats:
62% (01:49) correct
37% (02:16) wrong based on 67 sessions
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The center of a circle is (10, –3). The point (10, 9) is outside the circle, and the point (6, –3) is inside the circle; neither point is on the circle. If the radius, r, is an integer, how many possible values are there for r?
(A) Seven
(B) Eight
(C) Nine
(D) Ten
(E) Eleven
(A) Seven
(B) Eight
(C) Nine
(D) Ten
(E) Eleven
Re: The center of a circle is (10, –3). The point (10, 9) is out
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15 Nov 2018, 00:24
15 Nov 2018, 00:24
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Can I get an explanation?
Re: The center of a circle is (10, –3). The point (10, 9) is out
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19 Nov 2018, 14:11
19 Nov 2018, 14:11
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Explanation
Given that (10,9 ) is out side the circle centered at (10, –3). A distance between these two points are:
\(d1 = \sqrt{(10-10)^2+(9-(-3))^2}=12\). So the radius is less than 12.
Similarly, given that (6,-3) is out side the circle centered at (10, –3). A distance between these two points are:
\(d2 = \sqrt{(10-6)^2+(-3-(-3))^2}=4\). So the radius is more than 4.
So the radius is an integer between 4 and 12 hence there are 7 integers between 4 and 12.
Hence A.
Given that (10,9 ) is out side the circle centered at (10, –3). A distance between these two points are:
\(d1 = \sqrt{(10-10)^2+(9-(-3))^2}=12\). So the radius is less than 12.
Similarly, given that (6,-3) is out side the circle centered at (10, –3). A distance between these two points are:
\(d2 = \sqrt{(10-6)^2+(-3-(-3))^2}=4\). So the radius is more than 4.
So the radius is an integer between 4 and 12 hence there are 7 integers between 4 and 12.
Hence A.
