The radius of circle $x^{2} + y^{2} - 4 x - 6 y = 0$

\sqrt{13}

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\sqrt{10}

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\sqrt{2}

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\sqrt{24}

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Solution

The correct option is A $\sqrt{13}$
The equation of circle is $x^{2} + y^{2} - 4 x - 6 y = 0$
The radius of circle is $\sqrt{g^{2} + f^{2} - c} \sqrt{2^{2} + 3^{2} - 0} \sqrt{13}$

Suggest Corrections

Similar questions

x^{2} + y^{2} + 4 x + 6 y + 13 = 0

x^{2} + y^{2} + 4 x - 6 y - 12 = 0

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

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

2

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

2^{nd}

4^{th}

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