Question

Find the equation of the circle passing through the points (1,1) & (2,2) and whose radius is 1.

Answer 1

Firstly, we will have the required general equation of circle as:

(x – h)2 + (y – k)2 = r 2.

Now, it is given that the circle passes through the points (1, 1) and (2, 2) and radius is 1 unit. So, we will put the value of (x, y) = (1, 1) and (2, 2) one by one as:

(1 – h)2 + (1 – k)2 = 1 2 … (1)

(2 – h)2 + (2 – k)2 = 1 2 … (2)

On equating (1) and (2), we obtain

(1 – h)2 + (1 – k)2 = (2 – h)2 + (2 – k)2

⇒ 1 – 2h + h 2 + 1 – 2k + k 2 = 4 – 4h + h 2 + 4 – 4k + k 2

⇒ 2 – 2h – 2k = 8 – 4h – 4k

⇒ 2h + 2k = 6

⇒ h + k = 3 … (3)

⇒ h = 3 - k … (4)

On putting the value of h in (1), we get

(1 – (3 - k))2 + (1 – k)2 = 1

⇒(k - 2)2 + (1 – k)2 = 1

⇒ k2 + 4 - 4k + 1 + k2 - 2k = 1

⇒ 2k2 - 6k + 4 = 0

⇒ k2 - 3k + 2 = 0

⇒ (k - 2)(k - 1) = 0

⇒ k = 2 or k = 1

For k = 2 , we have h = 3 - 2 = 1

and for k = 1 , we have h = 3 - 1 = 2

On putting the two different set of values of h and k in the general equation, we get

For (h, k) = (2, 1):

(x – 2)2 + (y – 1)2 = 12

⇒(x – 2)2 + (y – 1)2 = 1 ... (5)

For (h, k) = (1, 2):

(x – 1)2 + (y – 2)2 = 12

⇒(x – 1)2 + (y – 2)2 = 1 .... (6)

Hence, equation (5) and (6) represent the required equation of circle.