Q. 124.5( 4 Votes )

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

Answer :

Given that we need to find the equation of the circle which passes through the points (1,1), (2,2) and having radius 1.

Let us assume the equation of the circle is:

⇒ x^{2} + y^{2} + 2ax + 2by + c = 0 .....(1)

We know that the radius of the circle is

⇒

⇒ a^{2} + b^{2} - c = 1 .....(2)

Substituting the point (1,1) in (1) we get,

⇒ 1^{2} + 1^{2} + 2a(1) + 2b(1) + c = 0

⇒ 1 + 1 + 2a + 2b + c = 0

⇒ 2a + 2b + c + 2 = 0

⇒ ..... (3)

Substituting the point (2,2) in (1) we get,

⇒ 2^{2} + 2^{2} + 2a(2) + 2b(2) + c = 0

⇒ 4 + 4 + 4a + 4b + c = 0

⇒ 4a + 4b + c + 8 = 0

⇒ ..... (4)

Subtracting (3) from (4), we get

⇒

⇒ c = 4 ..... (5)

Substituting (5) in (3) we get

⇒

⇒ a + b + 2 = - 1

⇒ a + b = - 3

⇒ a = - 3 - b ......(6)

Substituting (6) in (2) we get,

⇒ (- 3 - b)^{2} + b^{2} - 4 = 1

⇒ 9 + b^{2} + 6b + b^{2} - 5 = 0

⇒ 2b^{2} + 6b + 4 = 0

⇒ b^{2} + 3b + 2 = 0

⇒ b^{2} + 2b + b + 2 = 0

⇒ b(b + 2) + 1(b + 2) = 0

⇒ (b + 1)(b + 2) = 0

⇒ b + 1 = 0 (or) b + 2 = 0

⇒ b = - 1 (or) b = - 2

For b = - 1, substituting in (6)

⇒ a = - 3 - (- 1)

⇒ a = - 2

For b = - 2, substituting in (6)

⇒ a = - 3 - (- 2)

⇒ a = - 1

Now for a = - 2, b = - 1 and c = 4, the equation of the circle is x^{2} + y^{2} - 4x - 2x + 4 = 0

For a = - 1, b = - 2 and c = 4, the equation of the circle is x^{2} + y^{2} - 2x - 4y + 4 = 0

∴ The equations of the circles are x^{2} + y^{2} - 4x - 2y + 4 = 0 and x^{2} + y^{2} - 2x - 4y + 4 = 0.

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