Final answer:
To prove the two circles touch, we show that the condition for tangency, which is that the distance between centers equals the sum of radii, results in the given equation 1 /a² +1 /b²= 1 /c² after some algebraic manipulation. Thus, the condition is satisfied and the proof is complete.
Step-by-step explanation:
To prove that the two circles x² + y² + 2ax + c² = 0 and x² + y² + 2by + c² = 0 touch if 1 /a² +1 /b²= 1 /c², we need to understand the condition for two circles to touch each other.
Two circles will touch if the distance between their centers equals the sum of their radii. The first circle, with equation x² + y² + 2ax + c² = 0, has center (-a, 0) and radius √(a² - c²). The second circle, with equation x² + y² + 2by + c² = 0, has center (0, -b) and radius √(b² - c²).
The distance between their centers is √((-a-0)² + (0-(-b))²) = √(a² + b²). For the circles to touch, the sum of their radii must be equal to √(a² + b²). Therefore, √(a² - c²) + √(b² - c²) = √(a² + b²).
Squaring both sides of this equation gives a² - c² + 2√((a² - c²)(b² - c²)) + b² - c² = a² + b². This simplifies to 2√((a² - c²)(b² - c²)) = 2c². Dividing by 2 and then squaring both sides gives (a² - c²)(b² - c²) = c´, which leads to 1 /a² + 1/b² = 1/c² after dividing through by c´ and rearranging the terms. Thus, the condition is satisfied and the proof is complete.