If m_1, m_2 are the slopes of the tangents dr | vedclass

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(A) Given circle equation: $x^2+y^2-6x+4y+12=0$. Completing the square,we get $(x-3)^2+(y+2)^2=1$. This represents a circle with center $(3, -2)$ and

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$16$

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(A) Given circle equation: $x^2+y^2-6x+4y+12=0$. Completing the square,we get $(x-3)^2+(y+2)^2=1$. This represents a circle with center $(3, -2)$ and radius $r=1$. Let the slope of the tangent passing through $(1, -3)$ be $m$. The equation of the tangent is $y+3=m(x-1)$,which simplifies to $mx-y-m-3=0$. The perpendicular distance from the center $(3, -2)$ to the tangent must equal the radius $r=1$: $\left|\frac{m(3)-(-2)-m-3}{\sqrt{m^2+(-1)^2}}\right|=1$. $\left|\frac{2m-1}{\sqrt{m^2+1}}\right|=1$. Squaring both sides: $(2m-1)^2 = m^2+1$. $4m^2-4m+1 = m^2+1$. $3m^2-4m = 0$. $m(3m-4) = 0$. Thus,the slopes are $m_1=0$ and $m_2=\frac{4}{3}$. Finally,$9(m_1^2+m_2^2) = 9(0^2 + (\frac{4}{3})^2) = 9(\frac{16}{9}) = 16$.

If $m_1, m_2$ are the slopes of the tangents drawn from a point $(1, -3)$ to the circle $x^2+y^2-6x+4y+12=0$,then $9(m_1^2+m_2^2) = $

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