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

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

Vedclass · Accepted Answer

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

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(A) The center of the circle is the point of intersection of the diameters $2x - 3y = 5$ and $3x - 4y = 7$.Solving these equations:Multiply the first by $3$ and the second by $2$: $6x - 9y = 15$ and $6x - 8y = 14$.Subtracting the equations gives $y = -1$. Substituting $y = -1$ into $2x - 3(-1) = 5$ gives $2x + 3 = 5$,so $x = 1$.The center is $(1, -1)$.The area of the circle is $\pi r^2 = 154$.Using $\pi = \frac{22}{7}$,we have $\frac{22}{7} r^2 = 154$,which gives $r^2 = 154 	imes \frac{7}{22} = 49$,so $r = 7$.The equation of the circle is $(x - 1)^2 + (y + 1)^2 = 7^2$.Expanding this: $x^2 - 2x + 1 + y^2 + 2y + 1 = 49$.$x^2 + y^2 - 2x + 2y - 47 = 0$.

If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154 \text{ sq units}$,then the equation of the circle is (Take $\pi = \frac{22}{7}$)

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