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or the equation of the circle given by (x−5)2+y2=25, For the
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23 Feb 2020, 07:02
23 Feb 2020, 07:02
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For the equation of the circle given by (x−5)2+y2=25, For the equation of the circle given by (x−5)2+y2=25, find the minimum distance between the center of the circle and line y=−0.5x−10. (Choose the nearest approximation)
A. 9.5
B. 10.5
C. 11.5
D. 12.5
E. 13.5
A. 9.5
B. 10.5
C. 11.5
D. 12.5
E. 13.5
Re: or the equation of the circle given by (x−5)2+y2=25, For the
[#permalink]
23 Feb 2020, 07:58
23 Feb 2020, 07:58
from the circle equation, we can get center point of the circle
EX: (x-h)^2 + (y-k)^2 = r^2 center = (h,k)
so from (x-5)^2+y^2 = 25
we get center point = (5,0)
line equation = 0.5x+y+10
Formula for calculating distace, d= (|ax+by+c|)/sqrt(a^2+b^2)
sub x=5 and y=0
so, d= (0.5*5 + 10)/1.25
= 11.18
=11.5
EX: (x-h)^2 + (y-k)^2 = r^2 center = (h,k)
so from (x-5)^2+y^2 = 25
we get center point = (5,0)
line equation = 0.5x+y+10
Formula for calculating distace, d= (|ax+by+c|)/sqrt(a^2+b^2)
sub x=5 and y=0
so, d= (0.5*5 + 10)/1.25
= 11.18
=11.5
gmatclubot
Re: or the equation of the circle given by (x−5)2+y2=25, For the [#permalink]
23 Feb 2020, 07:58
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