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Question -

Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.



Answer -

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

Since the circle passesthrough 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 circlelies on line 4x + y = 16,

4h + k = 16 …(3)

From equations (1) and(2), we obtain

(4 – h)2 + (1 – k)= (6 – h)2 + (5 – k)2

16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2

16 – 8h + 1 –2k = 36 – 12h + 25 – 10k

4h + 8k =44

 h + 2k =11 … (4)

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

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

(4 – 3)2 + (1 – 4)2 = r2

(1)2 + (– 3)2 = r2

1 + 9 = r2

 r2 = 10

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

Since the circle passesthrough 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 circlelies on line 4x + y = 16,

4h + k = 16 …(3)

From equations (1) and(2), we obtain

(4 – h)2 + (1 – k)= (6 – h)2 + (5 – k)2

16 – 8h + h2 + 1 – 2k + k2 = 36 – 12h + h2 + 25 – 10k + k2

16 – 8h + 1 –2k = 36 – 12h + 25 – 10k

4h + 8k =44

 h + 2k =11 … (4)

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

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

(4 – 3)2 + (1 – 4)2 = r2

(1)2 + (– 3)2 = r2

1 + 9 = r2

 r2 = 10

Thus, the equation of therequired 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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