If $m_1, m_2$ are the slopes of the tangents drawn from a point (1, -3) to the circle ${\text{x}}^{\text{2}}{\text{\hspace{0.17em}+\hspace{0.17em}y}}^{\text{2}}\text{\hspace{0.17em}-\hspace{0.17em}6x\hspace{0.17em}+\hspace{0.17em}4y\hspace{0.17em}+\hspace{0.17em}12\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0}$ then  $9({\mathrm{m}}_1^2+{\mathrm{m}}_2^2)=$

Centre C = (3, -2), r = 1

Equation of tangent to circle in slope form 

is y + 2 = m(x – 3) $\pm \text{\hspace{0.17em}1}\sqrt{{\text{1+m}}^{\text{2}}}$ its passes 

through (1, -3) weget

3m² – 4m = 0

$\text{m\hspace{0.17em}=\hspace{0.17em}0\hspace{0.17em},\hspace{0.17em}\hspace{0.17em}}\frac{\text{4}}{\text{3}}$

$\text{ Let }{m}_1=0,m_2=\frac{4}{3}$

$\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}9}\left({\text{m}}_{\text{1}}^{\text{2}}{\text{\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}m}}_{\text{2}}^{\text{2}}\right)\text{\hspace{0.17em}=\hspace{0.17em}16}$