the center of a circle is (a+2 ,a-1).Find the value of a if the circle passes through [2,-2] and [8,-2]
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equations of circle =>r^2= (x-a-2)^2 + (y-a+1)^2
circle is passing through (2,-2)=
r^2 = (2-a-2)^2 + (-2-a+1)^2 = (-a)^2 + (-a-1)^2
circle is also passing through (8,-2) =
r^2 = (8-a-2)^2 + (-2-a+1)^2 = (6-a)^2 + (-a-1)^2
radius is same so
(-a)^2 + (-a-1)^2 = (6-a)^2 + (-a-1)^2
a^2 = 36 + a^2 - 12a
12a = 36
a = 3
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circle is passing through (2,-2)=
r^2 = (2-a-2)^2 + (-2-a+1)^2 = (-a)^2 + (-a-1)^2
circle is also passing through (8,-2) =
r^2 = (8-a-2)^2 + (-2-a+1)^2 = (6-a)^2 + (-a-1)^2
radius is same so
(-a)^2 + (-a-1)^2 = (6-a)^2 + (-a-1)^2
a^2 = 36 + a^2 - 12a
12a = 36
a = 3
PLEASE SELECT MY ANSWERS AS BRAINLIEST
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