Asked by Arya
A circle has its centre on x axis and passes through(5,1) And (3.4). Find its equation.
Answers
Answered by
Arora
Equation of a circle:
(x-h)^2 + (y-k)^2 = r^2
Where (x,y) is a point on the line and (h,k) is the center. If the center is on the x-axis, then k = 0
=> (x-h)^2 + y^2 = r^2
We've been given two points. In both, the radii will be the same, so,
(x1-h)^2 + (y1)^2 = (x2-h)^2 + (y2)^2
=> (5-h)^2 + 1^2 = (3-h)^2 + 4^2
=> 25 + h^2 - 10h + 1 = 9 + h^2 - 6h + 16
=> 1 = 4h
=> h = 1/4
Plugging this into the earlier equation, you get r^2 = 23.56
=> (x - 1.4)^2 + y^2 = 23.56
(x-h)^2 + (y-k)^2 = r^2
Where (x,y) is a point on the line and (h,k) is the center. If the center is on the x-axis, then k = 0
=> (x-h)^2 + y^2 = r^2
We've been given two points. In both, the radii will be the same, so,
(x1-h)^2 + (y1)^2 = (x2-h)^2 + (y2)^2
=> (5-h)^2 + 1^2 = (3-h)^2 + 4^2
=> 25 + h^2 - 10h + 1 = 9 + h^2 - 6h + 16
=> 1 = 4h
=> h = 1/4
Plugging this into the earlier equation, you get r^2 = 23.56
=> (x - 1.4)^2 + y^2 = 23.56
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