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30 Jun 2022, 21:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (02:09) correct
39%
(02:06)
wrong
based on 94
sessions
History
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What is the radius of the circle having centre at (4, 5) and passing through the centre of x^2 + y^2 + 4x - 6y = 2 ?
(A) 2√2
(B) 2√10
(C) 2√13
(D) 2√17
(E) 4√10
(A) 2√2
(B) 2√10
(C) 2√13
(D) 2√17
(E) 4√10
02 Jul 2022, 06:26
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Bunuel
Breaking Down the Info:
We would like to revert the equation into a standard circle form first by completing the square.
\(x^2 + y^2 + 4x - 6y = 2\)
\((x^2 + 4x + 4) + (y^2 - 6y + 9) = 2 + 4 + 9\)
\((x + 2)^2 + (y - 3)^2 = 15\). Then the center is (-2, 3).
The first circle has a center of (4, 5) and passes through (-2, 3), then the radius must be the distance between these two points.
The distance = \(\sqrt{6^2 + 2^2} = 2*\sqrt{3^2 + 1^2} = 2\sqrt{10}\)
Answer: B
