Question

If the circles x² + y² + 2ax + c = 0 and x² + y² + 2by + c = 0 touch each other, then

पर्याय

• \[\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c}\]

• \[\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}\]

• a + b = 2c

• \[\frac{1}{a} + \frac{1}{b} = \frac{2}{c}\]

Answer 1

\[\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c}\]

Given:
x² + y² + 2ax + c = 0    ...(1)
And, x² + y² + 2by + c = 0    ...(2)
For circle (1), we have:
Centre = \[\left( - a, 0 \right)\] = C₁ 

For circle (2), we have:
Centre = \[\left( 0, - b \right)\] = C₂

Let the circles intersect at point P.
∴ Coordinates of P = Mid point of C₁C₂
⇒ Coordinates of P = \[\left( \frac{- a + 0}{2}, \frac{0 - b}{2} \right) = \left( \frac{- a}{2}, \frac{- b}{2} \right)\]

Now, we have:

\[P C_1 = \text { radius of } \left( 1 \right)\]

\[ \Rightarrow \sqrt{\left( - a + \frac{a}{2} \right)^2 + \left( 0 - \frac{b}{2} \right)^2} = \sqrt{a^2 - c}\]

\[ \Rightarrow \frac{a}{4}^2 + \frac{b}{4}^2 = a^2 - c . . . \left( 3 \right)\]

\[\text { Also, radius of circle } \left( 1 \right) = \text { radius of circle } \left( 2 \right)\]

\[ \Rightarrow \sqrt{a^2 - c} = \sqrt{b^2 - c}\]

\[ \Rightarrow a^2 = b^2 . . . \left( 4 \right)\]

From (3) and (4), we have:

\[\frac{a^2}{2} = a^2 - c\]

\[ \Rightarrow \frac{a^2}{2} = c\]

\[ \Rightarrow \frac{2}{a^2} = \frac{1}{c}\]

\[ \Rightarrow \frac{1}{a^2} + \frac{1}{a^2} = \frac{1}{c}\]

\[ \Rightarrow \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c}\]