‌x2+y2−6x+4y+12=0 ⇒‌‌(x−3)2+(y+2)2=1 Equation of tangents ‌‌‌‌y+2=m(x−3)±1√1+m2 ∵‌ It passes through ‌(1,−3). ∴−3+2=m(1−3)±√1+m2⇒(2˙m−1)=±√1+m2 On squaring both sides, ‌4m2−4m+1=1+m2 ⇒‌‌3m2−4m=0⇒m(3m−4)=0 ⇒‌‌m=0,‌