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03 Jun 2020, 08:21
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The figure above shows a circle whose centre O has the x- and y- coordinates as (6,0). Points A(m,n) and B(8,2) are marked on the circle such that ∠AOB = 105∘. If the circle is symmetrical about the x-axis, what is the value of m?
A. 2
B. √6
C. 4
D. 6 - √6
E. Cannot be determined
Attachment:
1.png [ 19.44 KiB | Viewed 3217 times ]
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03 Jun 2020, 08:50
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Attachment:
1 (2).png [ 23.7 KiB | Viewed 3102 times ]
Angle AOY = 180-105-45=30
suppose radius is r
rcos45 = (8-6)
\(r=2√2\)
m=6 - rcos30 \(=6 - 2√2*\frac{√3}{2 }=6 -√6\)
03 Jun 2020, 21:28
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Another way is
Draw a perpendicular from B on OX and from A on OY.
1 (2).png [ 24.58 KiB | Viewed 3021 times ]
BC = 2; and OC = (8-6) = 2
Hence angle BOC = angle OBC = 45*
\(BO^2= BC^2+OC^2\)
\(BO^2 = 2^2+2^2 = 8\)
BO = 2√2
Angle AOD = 180* - 105* - 45* = 30*
Hence, triangle OAD is a (30-60-90) triangle.
AO = BO = 2√2 (both are radius of circle)
hence, \(OD = AO*\frac{√3}{2 }= √6 \)
m = 6 - OD = 6-√6
Draw a perpendicular from B on OX and from A on OY.
Attachment:
1 (2).png [ 24.58 KiB | Viewed 3021 times ]
BC = 2; and OC = (8-6) = 2
Hence angle BOC = angle OBC = 45*
\(BO^2= BC^2+OC^2\)
\(BO^2 = 2^2+2^2 = 8\)
BO = 2√2
Angle AOD = 180* - 105* - 45* = 30*
Hence, triangle OAD is a (30-60-90) triangle.
AO = BO = 2√2 (both are radius of circle)
hence, \(OD = AO*\frac{√3}{2 }= √6 \)
m = 6 - OD = 6-√6
