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 (2,3) and (-1,1).
(2 – h)2+ (3 – k)2 =r2 ……………..(1)
(-1 – h)2+ (1– k)2 =r2 ………………(2)
Since, the centre (h, k) of the circle lies on line x – 3y – 11= 0,
h – 3k =11………………… (3)
From the equation (1) and (2), we obtain
(2 – h)2+ (3 – k)2 =(-1 – h)2 + (1 – k)2
4 – 4h + h2 +9 -6k +k2 = 1 + 2h +h2+1 – 2k + k2
4 – 4h +9 -6k = 1 + 2h + 1 -2k
6h + 4k =11……………. (4)
Now let us multiply equation (3) by 6 and subtract it from equation (4) to get,
6h+ 4k – 6(h-3k) = 11 – 66
6h + 4k – 6h + 18k = 11 – 66
22 k = – 55
K = \( \dfrac{-5}{2}\)
Substitute this value of K in equation (4) to get,
6h + 4(-5/2) = 11
6h – 10 = 11
6h = 21
h = \( \dfrac{21}{6}\)
h = \( \dfrac{7}{2}\)
We obtain h =\( \dfrac{7}{2}\) and k = \( \dfrac{-5}{2}\)
On substituting the values of h and k in equation (1), we get
\( \dfrac{130}{4}\) = r2
The equation of the required circle is
4x2 -28x + 49 +4y2 + 20y + 25 =130
4x2 +4y2 -28x + 20y – 56 = 0
4(x2 +y2 -7x + 5y – 14) = 0
x2 + y2 – 7x + 5y – 14 = 0
The equation of the required circle is x2 + y2 – 7x + 5y – 14 = 0
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