Question

If the circles \( x^2 + y^2 - 2x + 4y + c = 0 \) and \( x^2 + y^2 + 2x - 4y + c = 0 \) have four common tangents, then

• A. \( c<0 \)
• B. \( -2<c<2 \)
• C. \( 0<c<5 \)
• D. \( c>0 \)

Hint:

Two circles have four common tangents if and only if they are externally separated. This condition is met when the distance between their centers is greater than the sum of their radii. Find the centers and radii of both circles. Calculate the distance between the centers and the sum of the radii. Apply the condition for externally separated circles to find the range of \( c \). Remember that for the radii to be real, the term under the square root must be positive.

Answer 1

The Correct Option is C

The equation of the first circle is \( S_1: x^2 + y^2 - 2x + 4y + c = 0 \).
Center \( C_1 = (1, -2) \), radius \( r_1 = \sqrt{(-1)^2 + (2)^2 - c} = \sqrt{1 + 4 - c} = \sqrt{5 - c} \).
The equation of the second circle is \( S_2: x^2 + y^2 + 2x - 4y + c = 0 \).
Center \( C_2 = (-1, 2) \), radius \( r_2 = \sqrt{(1)^2 + (-2)^2 - c} = \sqrt{1 + 4 - c} = \sqrt{5 - c} \).
For four common tangents, the circles must be externally separated, which means the distance between their centers must be greater than the sum of their radii.
Distance between centers \( d = \sqrt{(1 - (-1))^2 + (-2 - 2)^2} = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \).
Sum of radii \( r_1 + r_2 = \sqrt{5 - c} + \sqrt{5 - c} = 2\sqrt{5 - c} \).
For four common tangents, \( d>r_1 + r_2 \): \( 2\sqrt{5}>2\sqrt{5 - c} \) \( \sqrt{5}>\sqrt{5 - c} \) Squaring both sides: \( 5>5 - c \) \( c>0 \) Also, for the radii to be real, \( 5 - c>0 \implies c<5 \).
Combining these conditions, we get \( 0<c<5 \).