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Question
Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is ______.
Options
x2 + y2 + 13y = 0
3x2 + 3y2 + 13x + 3 = 0
6x2 + 6y2 – 13x = 0
x2 + y2 + 13x + 3 = 0
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Solution
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
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