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secondary 3 | A Maths
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I get the coordinates of center as (a,-a). But can’t find a. The ans is (6,-6). Radius is 6. Pls help
So, the diameter of the circle must be 12.
Its radius is 6.
With the circle being in the bottom right quadrant, its centre must be (6, -6).
Need me to draw the diagram?
Positive x-axis is where y = 0.
So if x = 0 and x = 12 are tangents to the circle, and both lines are parallel to each other, the centre is equidistant from these two lines.
Since the horizontal distance between x = 0 and x = 12 is 12 units,
The midpoint must be where x = 6
( (0 + 12)/2 )
This means x = 6 cuts through the centre of the circle and part of it is its diameter.
Now how to know the y-coordinate of the centre is -6?
Firstly, the circle lies below the x-axis and to the right of the y-axis since it is tangent to the negative y-axis and positive x-axis.
(We can say 4th quadrant)
Secondly, the diameter of the circle is now known to be 12. So the radius must be 6.
Since positive x-axis is tangent to the circle , and since y = 0 on this axis,
Then the highest point of this circle is at y = 0.
Shift 6 units down, we get y = -6. This is where the centre lies on.
Did the textbook steal the idea from there? Hmm...
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