Q. 144.0( 7 Votes )

# Find the equation of the circle which passes through the points A(1, 1) and B(2, 2) and whose radius is 1. Show that there are two such circles.

Answer :

The general equation of a circle: (x - h)^{2} + (y - k)^{2} = r^{2}

…(i), where (h, k) is the centre and r is the radius.

Putting A(1, 1) in (i)

(1 - h)^{2} + (1 - k)^{2} = 1^{2}

h^{2} + k^{2} + 2 - 2h - 2k = 1

h^{2} + k^{2} - 2h - 2k = - 1..(ii)

Putting B(2, 2) in (i)

(2 - h)^{2} + (2 - k)^{2} = 1^{2}

h^{2} + k^{2} + 8 - 4h - 4k = 1

h^{2} + k^{2} - 4h - 4k = - 7

(h^{2} + k^{2} - 2h - 2k) - 2h - 2k = - 7

- 1 - 2h - 2k = - 7 [from (ii)]

- 2h - 2k = - 6

h + k = 3 h= 3 - k

Putting it in (ii)

(3 - k)^{2} + k^{2} - 2(3 - k) - 2k = - 1

9 + 2k^{2} - 6k - 6 + 2k - 2k = - 1

2k^{2} + 4 - 6k = 0

k^{2} - 3k + 2 = 0

k = 2 or k = 1

When k = 2, h = 3 - 2 = 1

Equation of 1 circle

(x - 1)^{2} + (y - 2)^{2} = 1

When k = 1, h = 3 - 1 = 2

(x - 2)^{2} + (y - 1)^{2} = 1

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