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(A) The center of the circle is the point of intersection of the two diameters.Solving the system of equations:$2x - 3y = 5$  $(i)$$3x - 4y = 7$  $(ii

Vedclass · Accepted Answer

$x^{2} + y^{2} - 2x + 2y - 62 = 0$

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(A) The center of the circle is the point of intersection of the two diameters.Solving the system of equations:$2x - 3y = 5$  $(i)$$3x - 4y = 7$  $(ii)$Multiplying $(i)$ by $3$ and $(ii)$ by $2$:$6x - 9y = 15$$6x - 8y = 14$Subtracting the equations: $-y = 1 \Rightarrow y = -1$.Substituting $y = -1$ into $(i)$:$2x - 3(-1) = 5$ $\Rightarrow 2x + 3 = 5$ $\Rightarrow 2x = 2$ $\Rightarrow x = 1$.So,the center $(h, k)$ is $(1, -1)$ and the radius $r = 8$.The equation of the circle is $(x - h)^{2} + (y - k)^{2} = r^{2}$.$(x - 1)^{2} + (y - (-1))^{2} = 8^{2}$$(x - 1)^{2} + (y + 1)^{2} = 64$$x^{2} - 2x + 1 + y^{2} + 2y + 1 = 64$$x^{2} + y^{2} - 2x + 2y + 2 = 64$$x^{2} + y^{2} - 2x + 2y - 62 = 0$.

Find the equation of a circle whose diameters are $2x - 3y = 5$ and $3x - 4y = 7$ and whose radius is $8$.

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