If the area of the triangle formed by the pos | vedclass

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(B) The circle is $(x-2)^{2}+(y-3)^{2}=25$. The center is $(2,3)$ and the radius is $5$.At point $(5,7)$,the slope of the radius is $m_r = \frac{7-3}{

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$1225$

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(B) The circle is $(x-2)^{2}+(y-3)^{2}=25$. The center is $(2,3)$ and the radius is $5$.At point $(5,7)$,the slope of the radius is $m_r = \frac{7-3}{5-2} = \frac{4}{3}$.The slope of the tangent is $m_t = -\frac{1}{m_r} = -\frac{3}{4}$.The equation of the tangent is $y-7 = -\frac{3}{4}(x-5) \implies 4y-28 = -3x+15 \implies 3x+4y-43=0$.The $x-$intercept of the tangent is found by setting $y=0$: $3x = 43 \implies x = \frac{43}{3}$.The slope of the normal is $m_n = \frac{4}{3}$.The equation of the normal is $y-7 = \frac{4}{3}(x-5) \implies 3y-21 = 4x-20 \implies 4x-3y+1=0$.The $x-$intercept of the normal is found by setting $y=0$: $4x = -1 \implies x = -\frac{1}{4}$.The triangle is formed by the points $(-\frac{1}{4}, 0)$,$(\frac{43}{3}, 0)$,and $(5, 7)$.The base of the triangle on the $x-$axis is $b = \frac{43}{3} - (-\frac{1}{4}) = \frac{172+3}{12} = \frac{175}{12}$.The height of the triangle is the $y-$coordinate of the point $(5,7)$,which is $h = 7$.Area $A = \frac{1}{2} 	imes b 	imes h = \frac{1}{2} 	imes \frac{175}{12} 	imes 7 = \frac{1225}{24}$.Therefore,$24A = 1225$.

If the area of the triangle formed by the positive $x-$axis,the normal and the tangent to the circle $(x-2)^{2}+(y-3)^{2}=25$ at the point $(5,7)$ is $A$,then $24A$ is equal to ...... .

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