The equation of circle with centre at (2, -3) and radius 5 is

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

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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$
Equation of circle with centre (h.k) and radius r is :
$(x - h)^{2} + (y - k)^{2} = r^{2}$
Hence, equation of circle with centre (2,-3) and radius 5 is :
$(x - 2)^{2} + (y + 3)^{2} = 25$
$x^{2} + 4 - 4 x + y^{2} + 9 + 6 y = 25$
$x^{2} + y^{2} - 4 x + 6 y - 12 = 0$

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Similar questions

The diameters of a circle are along $2 x + y - 7 = 0$ and $x + 3 y - 11 = 0$ Then the equation of this circle, which also passes through (5,7) is

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

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

2^{nd}

4^{th}

$x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ and $x^{2} + y^{2} + 6 x + 18 y + 26 = 0$ __________.

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

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