The straight line 2x + 3y - k = 0, k 0 cuts | vedclass

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(A) Given the equation of the line: $2x + 3y - k = 0, k > 0$. Rewriting in intercept form: $\frac{x}{k/2} + \frac{y}{k/3} = 1$. Thus,the coordinates o

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

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(A) Given the equation of the line: $2x + 3y - k = 0, k > 0$. Rewriting in intercept form: $\frac{x}{k/2} + \frac{y}{k/3} = 1$. Thus,the coordinates of $A$ and $B$ are $(\frac{k}{2}, 0)$ and $(0, \frac{k}{3})$ respectively. The area of $	riangle OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{k}{2} 	imes \frac{k}{3} = \frac{k^{2}}{12}$. Given area $= 12$,so $\frac{k^{2}}{12} = 12$ $\Rightarrow k^{2} = 144$ $\Rightarrow k = 12$ (since $k > 0$). Therefore,$A = (6, 0)$ and $B = (0, 4)$. The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. Substituting the points $(6, 0)$ and $(0, 4)$: $(x - 6)(x - 0) + (y - 0)(y - 4) = 0$ $x(x - 6) + y(y - 4) = 0$ $x^{2} - 6x + y^{2} - 4y = 0$ $x^{2} + y^{2} - 6x - 4y = 0$.

The straight line $2x + 3y - k = 0, k > 0$ cuts the $x$ and $y$-axes at $A$ and $B$ respectively. The area of $\triangle OAB$,where $O$ is the origin,is $12 \text{ sq unit}$. The equation of the circle having $AB$ as diameter is

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