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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Rose34 on March 09, 2020, 07:05:15 pm
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Find the equation of the circle with centre (2,−3) which touches the x-axis.
My working out:
(x-h)^2 +(y-k)=r^2
(0-2)^2 + (0--3)^2= r^2
4+9=r^2
13=r^2
(x-2)^2 + (y+3)^2= 13
But apparently the correct answer is
(x−2)^2 + (y + 3)^2 = 9
Thanks in advance!!
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(0-2)^2 + (0--3)^2= r^2
By substituting in x=0 and y=0, you are assuming that the circle passes through (0, 0), but that is not given in the question.
Notice that, in general, if a circle just *touches* an axis (ie, does not cut through at two points), then the distance from the centre of the circle to that axis-intercept is equal to the radius.
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By substituting in x=0 and y=0, you are assuming that the circle passes through (0, 0), but that is not given in the question.
Notice that, in general, if a circle just *touches* an axis (ie, does not cut through at two points), then the distance from the centre of the circle to that axis-intercept is equal to the radius.
Thanks for the help I thought when it said that it *touches* it meant (0,0)