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The two points P and Q having coordinates (6, 0) and (0, 8) respectiv
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08 Dec 2024, 16:17
08 Dec 2024, 16:17
Expert Reply
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Question Stats:
80% (00:59) correct
20% (02:32) wrong based on 5 sessions
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The two points P and Q having coordinates (6, 0) and (0, 8) respectively, are situated at the two opposite ends of a circle C. The radius of circle C is
A. 10
B. 20
C. 5
D. 2
E. 4
A. 10
B. 20
C. 5
D. 2
E. 4
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The two points P and Q having coordinates (6, 0) and (0, 8) respectiv
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22 Dec 2024, 13:10
22 Dec 2024, 13:10
Expert Reply
The distance between two points $\(P(6,0) \& Q(0,8)\)$, which are situated at the two opposite ends of the circle C , will be the diameter of the circle.
So, the length of the diameter PQ is $\(\sqrt{(6-0)^2+(0-8)^2}=\sqrt{36+64}=\sqrt{100}=10\)$
$\( Distance between points\) $\(\left(x_1, y_1\right) \&\left(x_2, y_2\right)\)$ is $\(\left.\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \)$
Thus, the radius of the circle is $\(\frac{1}{2}($ Diameter $)=\frac{1}{2} \times 10=5\)$
Hence the answer is (C).
So, the length of the diameter PQ is $\(\sqrt{(6-0)^2+(0-8)^2}=\sqrt{36+64}=\sqrt{100}=10\)$
$\( Distance between points\) $\(\left(x_1, y_1\right) \&\left(x_2, y_2\right)\)$ is $\(\left.\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \)$
Thus, the radius of the circle is $\(\frac{1}{2}($ Diameter $)=\frac{1}{2} \times 10=5\)$
Hence the answer is (C).
gmatclubot
The two points P and Q having coordinates (6, 0) and (0, 8) respectiv [#permalink]
22 Dec 2024, 13:10
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