Question

Show that the circles $x^2+y^2-8x-2y+8=0$ and $x^2+y^2-2x+6y+6=0$ touch each other and find the point of contact

Answer 1

$x^2+y^2-8x-2y+8=0\Rightarrow {(x-4)}^2+{(y-1)}^2=9=3^2$

$x^2+y^2-2x+6y+6=0\Rightarrow {(x-1)}^2+{(y+3)}^2=4=2^2$

Distance between the centers $=\sqrt{{(4-1)}^2+{(1-(-3\left)\right)}^2}$

$=\sqrt{3^2+4^2}$

$=5$

$=3+2$

$r_1+r_2$

So, $|C_1{C}_2|=r_1+r_2\Rightarrow$ They touch each other externally.

P divides $C_1{C}_2$ in $3:2$ internally.

$P=\frac{3C_2+2C_1}{3+2}=\frac{3C_2+2C_1}{5}$

$=(\frac{3+8}{5},\frac{-9+2}{5})$

$=(\frac{11}{5},\frac{-7}{5})$