Equation of the circle passing through origin whose centre lie in the first quadrant and length of intercept on x and y-axis is $6$ and $4$ respectively, is -

x^{2} + y^{2} - 4 x - 6 y = 0

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x^{2} + y^{2} - 6 x - 4 y = 0

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x^{2} + y^{2} - 3 x - 2 y = 0

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None of these

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Solution

The correct option is C $x^{2} + y^{2} - 6 x - 4 y = 0$
Given circle is passing through $(0, 0), (6, 0) (0, 4)$
Hence using above formula required circle is $x^{2} + y^{2} - 6 x - 4 y = 0$

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Similar questions

The equation of the circle passing through the points (4,1),(6,5) whose centre lies on the line 4x+y-16=0 is

x^{2} + y^{2} - 4 x - 6 y - 3 = 0

x^{2} + y^{2} + 4 x - 2 y - 4 = 0

x^{2} + y^{2} - 6 x - 4 y = 0

2^{nd}

4^{th}

2 x - y - 2 = 0

(3, - 2)

(- 2, 0)

10

12

x

y

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