Asked by siddharth
Find the equation to the circle which has its centre at the point (3,4) and touch the straight line 5x 12y=1
Answers
Answered by
Reiny
You are missing an operator in your line equation, I will assume you mean
5x + 12y = 1 or 5x + 12y - 1 = 0
we know that the equation must be
(x-3)^2 + (y-4)^2 = r^2 , so the only thing missing is the value of r
using the distance from a point to a line method,
r = |5(3) + 12(4)-1 |/√(3^2 + 4^2) = 62/5
(x-3)^2 + (y-4)^2 = 3844/25
5x + 12y = 1 or 5x + 12y - 1 = 0
we know that the equation must be
(x-3)^2 + (y-4)^2 = r^2 , so the only thing missing is the value of r
using the distance from a point to a line method,
r = |5(3) + 12(4)-1 |/√(3^2 + 4^2) = 62/5
(x-3)^2 + (y-4)^2 = 3844/25
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