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(A) Given circles are $x^2+y^2+6x-8y+16=0$ and $x^2+y^2-2x-2y+1=0$. Converting to standard form $(x-h)^2+(y-k)^2=r^2$: $(x+3)^2+(y-4)^2=3^2$ and $(x-1

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$\frac{15}{4}$

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(A) Given circles are $x^2+y^2+6x-8y+16=0$ and $x^2+y^2-2x-2y+1=0$. Converting to standard form $(x-h)^2+(y-k)^2=r^2$: $(x+3)^2+(y-4)^2=3^2$ and $(x-1)^2+(y-1)^2=1^2$. Thus,$C_1=(-3,4), r_1=3$ and $C_2=(1,1), r_2=1$. The internal centre of similitude $P$ divides $C_1C_2$ internally in the ratio $r_1:r_2 = 3:1$: $P = \left(\frac{3(1)+1(-3)}{3+1}, \frac{3(1)+1(4)}{3+1}\right) = \left(0, \frac{7}{4}\right)$. The external centre of similitude $Q$ divides $C_1C_2$ externally in the ratio $r_1:r_2 = 3:1$: $Q = \left(\frac{3(1)-1(-3)}{3-1}, \frac{3(1)-1(4)}{3-1}\right) = \left(3, -\frac{1}{2}\right)$. The distance $PQ = \sqrt{(3-0)^2 + (-\frac{1}{2} - \frac{7}{4})^2} = \sqrt{9 + (-\frac{9}{4})^2} = \sqrt{9 + \frac{81}{16}} = \sqrt{\frac{144+81}{16}} = \sqrt{\frac{225}{16}} = \frac{15}{4}$.

The distance between the centres of similitude of the circles $x^2+y^2+6x-8y+16=0$ and $x^2+y^2-2x-2y+1=0$ is

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