Let the circle f(x,y)=x2+y2+10x−12y+51=0 and the line g(x,y)=3y+x−3=0 intersect at a point P(a, b). ∴ g(a, b) = 0 3b + a - 3 = 0 ⇒ a = 3 - 3b ... (1) And, f(a, b) = 0 ⇒a2+b2+10a−12b+51=0 Using equation (1), we get: ⇒(3−3b)2+b2+10(3−3b)−12b+51=0 ⇒9−18b+9b2+b2+30−30b−12b+51=0 ⇒10b2−60b+90=0 ⇒b2−6b+9=0 ⇒(b−3)2=0 ⇒b=3 And, using equation (1): a=3−3×3=3−9=−6. Therefore, the given curves intersect at point (-6, 3).