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Question

Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.


Solution

Let the equation of the required circle be (xh)2 + (yk)2 = r2.

Since 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 equations (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)

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

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

Thus, the equation of the required circle is


Mathematics
Part - I
Standard XII

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