Consider the circle x^2+y^2-6x+4y=12. The equ | vedclass

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(D) The given equation of the circle is $x^2+y^2-6x+4y=12$. Completing the square,we get $(x-3)^2+(y+2)^2 = 12+9+4 = 25 = 5^2$. The center is $(3, -2)

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$4x+3y-31=0$

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(D) The given equation of the circle is $x^2+y^2-6x+4y=12$. Completing the square,we get $(x-3)^2+(y+2)^2 = 12+9+4 = 25 = 5^2$. The center is $(3, -2)$ and the radius $r=5$. $A$ line parallel to $4x+3y+5=0$ is of the form $4x+3y+k=0$. The distance from the center $(3, -2)$ to the tangent line $4x+3y+k=0$ must be equal to the radius $r=5$. Using the distance formula,$\frac{|4(3)+3(-2)+k|}{\sqrt{4^2+3^2}} = 5$. $\frac{|12-6+k|}{5} = 5 \Rightarrow |6+k| = 25$. This gives $6+k = 25$ or $6+k = -25$. So,$k = 19$ or $k = -31$. The equations of the tangents are $4x+3y+19=0$ and $4x+3y-31=0$.

Consider the circle $x^2+y^2-6x+4y=12$. The equations of a tangent to this circle that is parallel to the line $4x+3y+5=0$ are

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