Question

Prove that the circles $x^2+y^2-14x-10y+58=0$ and $x^2+y^2-2x+6y-26=0$ touch each other

Answer 1

$x^2+y^2-14x-10y+58=0$

$\Rightarrow (x-7{)}^2+(y-5{)}^2=16$

So this circle has center $C_1=(7,5)$ and $R_1=4$

$x^2+y^2-2x+6y-26=0$

$\Rightarrow (x-1{)}^2+(y+3{)}^2=36$

So this circle has center $C_2=(1,-3)$ and $R_2=6$

Therefore,

$C_1{C}_2=\sqrt{(7-1{)}^2+(5+3{)}^2}=10$ and $R_1+R_2=4+6=10$

Since the distance between the centers is equal to the sum of the radii, these circles touch each other externally.