The distance between the centres of similitude of the circles $x^2+y^2+6 x-8 y+16=0$ and $x^2+y^2-2 x-2 y+$…
The distance between the centres of similitude of the circles $x^2+y^2+6 x-8 y+16=0$ and $x^2+y^2-2 x-2 y+$ $1=0$ is
$\frac{15}{4}$
$\frac{5}{4}$
$\frac{5}{2}$
$\frac{15}{2}$
Solution
Given circles $x^2+y^2+6 x-8 y+16=0$
and $x^2+y^2-2 x-2 y+1=0$
$\Rightarrow(x+3)^2+(y-4)^2=3^2,(x-1)^2+(y-1)^2=1^2$
So $C_1=(-3,4), C_2=(1,1), r_1=3 \& r_2=1$
Let internal centre $P=\left(\frac{3(1,1)+1(-3,4)}{1+3}\right)$
$=\left(0, \frac{7}{4}\right)$
external centre $Q=\left(\frac{3(1,1)-1(-3,4)}{3-1}\right)$
$=\left(\frac{6}{2}, \frac{-1}{2}\right)=\left(3, \frac{-1}{2}\right)$
Now $P Q=\sqrt{(0-3)^2+\left(\frac{7}{4}+\frac{1}{2}\right)^2}$
$=\sqrt{9+\frac{81}{16}}=\sqrt{\frac{225}{16}}=\frac{15}{4}$