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