Vì (d)//3x+4y-5=0 nên (d): 3x+4y+c=0
(C): \(\left(x-1\right)^2+\left(y-2\right)^2=25\)
=>Tâm là I(1;2); R=5
(d) tiếp xúc với (C) nên \(d\left(I;\left(d\right)\right)=R=5\)
=>\(\dfrac{\left|3\cdot1+4\cdot2+c\right|}{\sqrt{3^2+4^2}}=5\)
=>\(\left|c+11\right|=25\)
=>\(\left[{}\begin{matrix}c+11=25\\c+11=-25\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}c=14\\c=-36\end{matrix}\right.\)
Khi c=14 thì (d) có dạng là \(3x+4y+14=0\)
=>\(x+\dfrac{4}{3}y+\dfrac{14}{3}=0\)
=>\(a=\dfrac{4}{3};b=\dfrac{14}{3}\)
=>\(a+b=\dfrac{18}{3}=6\)
Khi c=-36 thì (d): \(3x+4y-36=0\)
=>\(x+\dfrac{4}{3}y-12=0\)
=>\(a=\dfrac{4}{3};b=-12\)
=>\(a+b=\dfrac{4}{3}-12=\dfrac{4}{3}-\dfrac{36}{12}=-\dfrac{32}{12}\)
