Let us consider the equation of the required circle be
(x – h)2+ (y – k)2 = r2
We know that the circle passes through points (4,1) and (6,5)
So,
(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 =\( \sqrt{10}\)
so now, (x – 3)2 + (y – 4)2 = \( (\sqrt{10})^2\)
x2 – 6x + 9 + y2 – 8y + 16 =10
x2 + y2 – 6x – 8y + 15 = 0
The equation of the required circle is x2 + y2 – 6x – 8y + 15 = 0
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