Q.(a) Show that the circles $x^2+y^2-2x+6y+6=0$ and $x^2+y^2-5x+6y+15=0$ touch each other. OR

Find each circle's centre and radius, compute the distance between the centres, and show it equals $|r_1-r_2|$, which means the circles touch internally.

For a circle $x^2+y^2+2gx+2fy+c=0$, the centre is $(-g,-f)$ and the radius is $\sqrt{g^2+f^2-c}$.

Circle 1: $x^2+y^2-2x+6y+6=0 \Rightarrow 2g=-2,\ 2f=6,\ c=6 \Rightarrow g=-1,\ f=3$.

Centre $C_1=(1,-3)$, radius $r_1=\sqrt{(-1)^2+3^2-6}=\sqrt{1+9-6}=\sqrt{4}=2$.

Circle 2: $x^2+y^2-5x+6y+15=0 \Rightarrow 2g=-5,\ 2f=6,\ c=15 \Rightarrow g=-2.5,\ f=3$.

Centre $C_2=(2.5,-3)$, radius $r_2=\sqrt{(-2.5)^2+3^2-15}=\sqrt{6.25+9-15}=\sqrt{0.25}=0.5$.

Distance between centres:

$C_1C_2 = \sqrt{(2.5-1)^2+(-3-(-3))^2} = \sqrt{(1.5)^2+0} = 1.5$

Compare with the radii:

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