The equation of circle x2+y2+4x−6y=12 can be written as (x+2)2+(y−3)2=25
Let P=(1,−1)&Q=(4,0) Equation of tangent at P(1,−1) to the given circle : x(1)+y(−1)+2(x+1)−3(y−1)−12=0 3x−4y−7=0 . . . (i) The required circle is tangent to (1) at (1,-1) ∴(x−1)2+(y+1)2+λ(3x−4y−7)=0 . . . (ii) Equation (ii) passes through Q(4,0) ⇒32+12+λ(12−7)=0⇒5λ+10=0⇒λ=−2 Equation (2) becomes x2+y2−8x+10y+16=0 radius =√(−4)2+(5)2−16=5