# Find the eq

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

Since, the circle passes through points (4,1) and (6,5),

(4 – h)2+ (1 – k)2 =r2 .................(1)

(6– h)2+ (5 – k)2 =r2 ..................(2)

Since, the centre (h,k) of the circle lies on line 4x+y = 16,

4h + k =16..................... (3)

From the equation (1) and (2), we obtain

(4 – h)2+ (1 – k)2 =(6 – h)2 + (5 – k)2 16 – 8h + h2 +1 -2k +k2 = 36 -12h +h2+15 – 10k + k2 16 – 8h +1 -2k + 12h -25 -10k 4h +8k = 44 h + 2k =11................ (4)

On solving equations (3) and (4), we obtain h=3 and k= 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)2+ (1 – 4)2 =r2 (1)2 + (-3)2 = r2 1+9 = r2 r = Thus, the equation of the requires circle is

(x – 3)2+ (y – 4)2 = x2 -6x + 9 + y2 – 8y + 16 =10

⇒ x2 + y2 -6x – 8y = 15 =0

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