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find the equation of a circle of radius 5 and the centre lies on the positive side of x axis and a distance of 5 unit from the origin
Hi Rahul,
It will be a family of circles satisfying the given conditions.
let ( x1, y1 ) be the centre of circle.
centre is at distance of 5 unit from origin , it means root(x12 + y12 ) = 5
thus x12 + y12 = 25
Eqn of family of circles will be ( X-x1)2 + ( Y - y1 )2 = 52 and x12 + y12 = 25
You can put different values of x1 and y1 ( for ex : 4,3 ) to get the equations of circles.
like (x - 4)2 + ( y - 3 )2 = 52
Please feel free to ask as many questions you have.
Puneet
let the center of circle be (h,k)
distance from origin ={(h-o)^2 + (k-0)^2}^1/2=5
point lies on x axis so k=0 and h=+_5
center of circle is (5,0)
equation of circle is (x-5)^2 +y^2=25
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