Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).


Let the equation of the required circle be (x – h)2+ (y – k)2 =r2


Since, the radius of the circle is 5 and its centre lies on the x- axis, k= 0 and r= 5.


Now, the equation of the circle becomes (x – h)2 + y2 = 25.


It is given that the circle passes through point (2, 3)


(2 – h)2+ 32 = 2


(2 – h)2 = 25-9


(2 – h)2 = 16


2 –h == 4


If 2-h =4, then h = -2


If 2-h = -4, then h =6.


When h= -2, the equation of the circle becomes


(x – 2)2 + y2 = 25


x2 -12x + 36 + y2 = 25


x2 + y2 -4x + 21 =0


When h= 6, the equation of the circle becomes


(x – 6)2 + y2 = 25


x2 -12x + 36 + y2 = 25


x2 + y2 -12x + 11 =0


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