If the line 3x - 2y + 6 = 0 meets the X-axis | vedclass

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(B) The line is given by $3x - 2y + 6 = 0$. To find point $A$ (where it meets the $X$-axis),set $y = 0$: $3x + 6 = 0 \Rightarrow x = -2$. So,$A = (-2,

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$x^2 + y^2 + 4x - 9 = 0$

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(B) The line is given by $3x - 2y + 6 = 0$. To find point $A$ (where it meets the $X$-axis),set $y = 0$: $3x + 6 = 0 \Rightarrow x = -2$. So,$A = (-2, 0)$. To find point $B$ (where it meets the $Y$-axis),set $x = 0$: $-2y + 6 = 0 \Rightarrow y = 3$. So,$B = (0, 3)$. The radius $r$ is the distance $AB = \sqrt{(0 - (-2))^2 + (3 - 0)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$. The equation of a circle with center $(h, k) = (-2, 0)$ and radius $r = \sqrt{13}$ is $(x - h)^2 + (y - k)^2 = r^2$. $(x - (-2))^2 + (y - 0)^2 = (\sqrt{13})^2$ $(x + 2)^2 + y^2 = 13$ $x^2 + 4x + 4 + y^2 = 13$ $x^2 + y^2 + 4x - 9 = 0$.

If the line $3x - 2y + 6 = 0$ meets the $X$-axis and $Y$-axis at points $A$ and $B$ respectively,then the equation of the circle with radius $AB$ and center at $A$ is:

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