Question

The circles whose equations are \(x^2 + y^2 + c^2 = 2ax\) and \(x^2 + y^2 + c^2 - 2by = 0\) will touch each other externally if

• A. \(\frac{1}{b^2} + \frac{1}{c^2} = \frac{1}{a^2}\)
• B. \(\frac{1}{c^2} + \frac{1}{a^2} = \frac{1}{b^2}\)
• C. \(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}\)
• D. None of these

Hint:

Use distance formula and condition for external tangency.

Answer 1

The Correct Option is C

To determine the condition for two circles to touch each other externally, we need to consider the given equations of the circles and match them with the standard equation of a circle.

1. The equation of the first circle is given as \(x^2 + y^2 + c^2 = 2ax\). This can be rewritten in standard form: \((x - a)^2 + y^2 = a^2 - c^2\).
2. The center of the first circle is \((a, 0)\), and the radius is \(\sqrt{a^2 - c^2}\).
3. The equation of the second circle is \(x^2 + y^2 + c^2 - 2by = 0\), which can be rewritten as \((x - 0)^2 + (y - b)^2 = b^2 - c^2\).
4. The center of the second circle is \((0, b)\), and the radius is \(\sqrt{b^2 - c^2}\).

For the circles to touch each other externally, the distance between their centers must be equal to the sum of their radii.

• The distance between the centers \((a, 0) \text{ and } (0, b)\) is \(\sqrt{a^2 + b^2}\).
• The sum of the radii is \(\sqrt{a^2 - c^2} + \sqrt{b^2 - c^2}\).

Equating the distance and the sum of the radii gives: \(\sqrt{a^2 + b^2} = \sqrt{a^2 - c^2} + \sqrt{b^2 - c^2}\).

Squaring both sides to eliminate the square roots:

• \(a^2 + b^2 = (a^2 - c^2) + (b^2 - c^2) + 2\sqrt{(a^2 - c^2)(b^2 - c^2)}\)

Solving this equation leads to:

• \(2c^2 = 2\sqrt{(a^2 - c^2)(b^2 - c^2)}\)

Squaring again and simplifying, we ultimately find the condition: \(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}\).

Thus, the circles will touch each other externally if \(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}\).