The equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is
(a) x² + y² + 13y = 0
(b) 3x² + 3y² + 13x + 3 = 0
(c) 6x² + 6y² – 13x = 0
(d) x² + y² + 13x + 3 = 0

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Solution

Circle with center as y-axis is of the form
x²+ (y − b)² = r² ...(1)
also given, circle passes through origin
⇒ 0² + (0 – b²) = r²
i.e r² = b²
i.e. equation of circle reduces to,
x² + (y – b)² = b²
i.e. x² + y² – 2by + b² = b²
i.e. x² + y² – 2by = 0
Given circle passes through (2 , 3)
⇒ 4 + 9 – 2b(3) = 0
i.e. 6b = 13
i.e. b = 13/6
∴ Equation of circle is x² + y² – 2 (13/6)y = 0
i.e. 3x² + 3y² – 13y = 0

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The equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is (a) x2 + y2 + 13y = 0 (b) 3x2 + 3y2 + 13x + 3 = 0 (c) 6x2 + 6y2 – 13x = 0 (d) x2 + y2 + 13x + 3 = 0

The equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is
(a) x² + y² + 13y = 0
(b) 3x² + 3y² + 13x + 3 = 0
(c) 6x² + 6y² – 13x = 0
(d) x² + y² + 13x + 3 = 0