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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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