The equation of the circle with centre (2, -3 | vedclass

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Question

(C) The centre of the circle is $(h, k) = (2, -3)$.Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coor

Vedclass · Accepted Answer

$x^2+y^2-4x+6y+4=0$

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(C) The centre of the circle is $(h, k) = (2, -3)$.Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the centre.$r = |k| = |-3| = 3$.The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 2)^2 + (y - (-3))^2 = 3^2$.$(x - 2)^2 + (y + 3)^2 = 9$.Expanding the terms: $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 9$.$x^2 + y^2 - 4x + 6y + 4 + 9 = 9$.$x^2 + y^2 - 4x + 6y + 4 = 0$.

The equation of the circle with centre $(2, -3)$ and touching the $X$-axis is

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