Question

Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is ______.

विकल्प

• x² + y² + 13y = 0

• 3x² + 3y² + 13x + 3 = 0

• 6x² + 6y² – 13x = 0

• x² + y² + 13x + 3 = 0

Answer 1

Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is x² + y² + 13y = 0.

Explanation:

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

Let the centre be (0, a)

∴ Radius r = a

So, the equation of the circle is (x – 0)² + (y – a)² = a²

⇒ x² + (y – a)² = a²

⇒ x² + y² + a² – 2ay = a²

⇒ x² + y² – 2ay = 0   ......(i)

Now CP = r

⇒ `sqrt((2 - 0)^2 + (3 - a)^2) = a`

⇒ `sqrt(4 + 9 + a^2 - 6a) = a`

⇒ `sqrt(13 + a^2 - 6a) = a`

⇒ `13 + a^2 - 6a = a^2`

⇒ `13 - 6a = 0`

∴ `a = 13/6`

Putting the value of a in equation (i) we get

`x^2 + y^2 - 2(13/6)y` = 0

⇒ 3x² + 3y² – 13y = 0