If 2x - 3y + 5 = 0 and 4x - 5y + 7 = 0 are th | vedclass

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(B) The center of the circle $(h, k)$ is the intersection point of the two normals.Solving the system of equations:$2x - 3y = -5$ $(1)$$4x - 5y = -7$

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$2$

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(B) The center of the circle $(h, k)$ is the intersection point of the two normals.Solving the system of equations:$2x - 3y = -5$ $(1)$$4x - 5y = -7$ $(2)$Multiply $(1)$ by $2$: $4x - 6y = -10$ $(3)$Subtract $(3)$ from $(2)$: $(4x - 5y) - (4x - 6y) = -7 - (-10) \implies y = 3$.Substitute $y = 3$ into $(1)$: $2x - 3(3) = -5 \implies 2x - 9 = -5 \implies 2x = 4 \implies x = 2$.So,the center of the circle is $(2, 3)$.The point $(2, 5)$ lies on the circle.The radius $r$ is the distance between the center $(2, 3)$ and the point $(2, 5)$.$r = \sqrt{(2 - 2)^2 + (5 - 3)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2$.

If $2x - 3y + 5 = 0$ and $4x - 5y + 7 = 0$ are the equations of the normals drawn to a circle and $(2, 5)$ is a point on the given circle,then the radius of the circle is

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