Asked by Cad
Find the equation of a circle with center C(1,6) and tangent line x-y-1=0.
Answers
Answered by
Reiny
equation of circle:
(x-1)^2 + (y-6)^2 = r^2
tangent line is y = x-1
When intersected , it should give us one root
(x-1)^2+ (x-1 - 6)^2 = r^2
x^2 - 2x + 1 + x^2 -14x + 47 - r^2 = 0
2x^2 - 16x + 48-r^2
a quadratic where
a = 2
b = -16
c = 48-r^2
to have only one root, b^2 - 4ac=0
256 - 4(2)(48-r^2) = 0
256 - 384 + 8r^2 = 0
r^2 = 16
(x-1)^2 + (y-6)^2 = 16
easier way:
distance from (1,6) to x-y-1 = 0 is
| (1-6-1) |/√(1+1) = 6/√2
r = 6/√2
r^2 = 36/2 = 16
equation: (x-1)^2 + (y-6)^2 = 16
(x-1)^2 + (y-6)^2 = r^2
tangent line is y = x-1
When intersected , it should give us one root
(x-1)^2+ (x-1 - 6)^2 = r^2
x^2 - 2x + 1 + x^2 -14x + 47 - r^2 = 0
2x^2 - 16x + 48-r^2
a quadratic where
a = 2
b = -16
c = 48-r^2
to have only one root, b^2 - 4ac=0
256 - 4(2)(48-r^2) = 0
256 - 384 + 8r^2 = 0
r^2 = 16
(x-1)^2 + (y-6)^2 = 16
easier way:
distance from (1,6) to x-y-1 = 0 is
| (1-6-1) |/√(1+1) = 6/√2
r = 6/√2
r^2 = 36/2 = 16
equation: (x-1)^2 + (y-6)^2 = 16
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