Question

Show that the circles
x^2 + y^2 - 14x - 10y + 58 = 0
and
x^2 + y^2 - 2x + 6y - 26 = 0
touch each other externally.​

Answer 1

Step-by-step explanation:

${x }^{2} + {y}^{2} - 14x - 10y = - 58 \: \: \: eq.(1) \\ {x}^{2} + {y}^{2} - 2x + 6y = 26 \: \: \: \: eq.(2)$

Eq.(1) ÷(2)

Answer 2

Answer:

This is the answer of your question.