Question

If \( (h, k) \) is the internal center of similitude of the circles \( x^2 + y^2 + 2x - 6y + 1 = 0 \) and \( x^2 + y^2 - 4x + 2y + 4 = 0 \), then find \( 4h \).

• A. 0
• B. 3
• C. 1
• D. 5 \bigskip

Hint:

The center of similitude is the point where the lines joining the centers of two circles meet.

Answer 1

The Correct Option is D

We are given two circles: \[ C_1: x^2 + y^2 + 2x - 6y + 1 = 0 \] \[ C_2: x^2 + y^2 - 4x + 2y + 4 = 0 \] We need to find \( 4h \), where \( (h, k) \) is the internal center of similitude of these circles. --- Step 1: Convert Circles to Standard Form The general equation of a circle: \[ x^2 + y^2 + Dx + Ey + F = 0 \] has center \( (-D/2, -E/2) \) and radius: \[ r = \sqrt{\left(\frac{D}{2}\right)^2 + \left(\frac{E}{2}\right)^2 - F} \] # Circle 1: \[ x^2 + y^2 + 2x - 6y + 1 = 0 \] - Center \( C_1 = (-1, 3) \) - Radius: \[ r_1 = \sqrt{\left(\frac{2}{2}\right)^2 + \left(\frac{-6}{2}\right)^2 - 1} \] \[ = \sqrt{1 + 9 - 1} = \sqrt{9} = 3 \] # Circle 2: \[ x^2 + y^2 - 4x + 2y + 4 = 0 \] - Center \( C_2 = (2, -1) \) - Radius: \[ r_2 = \sqrt{\left(\frac{-4}{2}\right)^2 + \left(\frac{2}{2}\right)^2 - 4} \] \[ = \sqrt{4 + 1 - 4} = \sqrt{1} = 1 \] --- Step 2: Internal Center of Similitude Formula The internal center of similitude is given by: \[ h = \frac{r_1 C_2x + r_2 C_1x}{r_1 + r_2}, \quad k = \frac{r_1 C_2y + r_2 C_1y}{r_1 + r_2} \] Substituting: \[ h = \frac{(3)(2) + (1)(-1)}{3 + 1} = \frac{6 - 1}{4} = \frac{5}{4} \] \[ k = \frac{(3)(-1) + (1)(3)}{3 + 1} = \frac{-3 + 3}{4} = \frac{0}{4} = 0 \] --- Step 3: Compute \( 4h \) \[ 4h = 4 \times \frac{5}{4} = 5 \] Thus, the correct answer is: \[ \boxed{5} \] \bigskip