If the lines 3x - 4y - 7 = 0 and 2x - 3y - 5 | vedclass

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(D) The center of the circle is the point of intersection of the two diameters $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$.Solving these equations: $3x -

Vedclass · Accepted Answer

$x^2 + y^2 - 2x + 2y - 47 = 0$

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(D) The center of the circle is the point of intersection of the two diameters $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$.Solving these equations: $3x - 4y = 7$ and $2x - 3y = 5$.Multiplying the first by $3$ and the second by $4$: $9x - 12y = 21$ and $8x - 12y = 20$.Subtracting gives $x = 1$. Substituting $x = 1$ into $2x - 3y = 5$ gives $2 - 3y = 5$,so $y = -1$.The center is $(1, -1)$.The area of the circle is $\pi r^2 = 49\pi$,so $r^2 = 49$ and $r = 7$.The equation of the circle is $(x - 1)^2 + (y + 1)^2 = 7^2$.Expanding this: $x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.Thus,$x^2 + y^2 - 2x + 2y - 47 = 0$.

If the lines $3x - 4y - 7 = 0$ and $2x - 3y - 5 = 0$ are two diameters of a circle of area $49\pi$ square units,the equation of the circle is:

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