The lines 2x - 3y = 5 and 3x - 4y = 7 are dia | vedclass

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

Vedclass · Accepted Answer

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

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(D) The center of the circle is the intersection point of the diameters $2x - 3y = 5$ and $3x - 4y = 7$.Solving the system: $2x - 3y = 5$ (multiplied by $3$) $\Rightarrow 6x - 9y = 15$ and $3x - 4y = 7$ (multiplied by $2$) $\Rightarrow 6x - 8y = 14$.Subtracting the equations: $(6x - 8y) - (6x - 9y) = 14 - 15 \Rightarrow y = -1$.Substituting $y = -1$ into $2x - 3(-1) = 5$ $\Rightarrow 2x + 3 = 5$ $\Rightarrow x = 1$.So,the center $(h, k) = (1, -1)$.Given the area $A = \pi r^2 = 154$. Taking $\pi \approx \frac{22}{7}$,we have $\frac{22}{7} r^2 = 154 \Rightarrow r^2 = 154 	imes \frac{7}{22} = 7 	imes 7 = 49$.Thus,$r = 7$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.$(x - 1)^2 + (y + 1)^2 = 49$.$x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.$x^2 + y^2 - 2x + 2y = 47$.

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are diameters of a circle with an area of $154$ square units. Find the equation of the circle.

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