Question

The distance between the centres of similitude of the circles \( x^2 + y^2 + 6x - 8y + 16 = 0 \) and \( x^2 + y^2 - 2x - 2y + 1 = 0 \) is:

• A. \( \frac{15}{4} \)
• B. \( \frac{5}{4} \)
• C. \( \frac{5}{2} \)
• D. \( \frac{15}{2} \)

Hint:

To find the distance between the centres of similitude, use the formula \( d = \frac{|r_1^2 - r_2^2|}{|r_1 - r_2|} \) and substitute the known values.

Answer 1

The Correct Option is A

We are given two circles: 1. \( x^2 + y^2 + 6x - 8y + 16 = 0 \) 2. \( x^2 + y^2 - 2x - 2y + 1 = 0 \) The equation of a circle is given by \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the centre of the circle and \( r \) is its radius. Step 1: Rewrite the equations of the circles in standard form For the first circle, complete the square: \[ x^2 + 6x + y^2 - 8y = -16 \] Complete the square for \( x \) and \( y \): \[ (x + 3)^2 + (y - 4)^2 = 9 \] Thus, the centre of the first circle is \( (-3, 4) \) and its radius is \( 3 \). For the second circle, complete the square: \[ x^2 - 2x + y^2 - 2y = -1 \] Complete the square for \( x \) and \( y \): \[ (x - 1)^2 + (y - 1)^2 = 3 \] Thus, the centre of the second circle is \( (1, 1) \) and its radius is \( \sqrt{3} \). Step 2: Find the distance between the centres of similitude The distance between the centres of similitude of two circles is given by the formula: \[ d = \frac{|r_1^2 - r_2^2|}{|r_1 - r_2|} \] where \( r_1 \) and \( r_2 \) are the radii of the two circles. For the first circle, \( r_1 = 3 \), and for the second circle, \( r_2 = \sqrt{3} \). Substitute into the formula: \[ d = \frac{|3^2 - (\sqrt{3})^2|}{|3 - \sqrt{3}|} = \frac{|9 - 3|}{|3 - \sqrt{3}|} = \frac{6}{|3 - \sqrt{3}|} \] We can approximate \( 3 - \sqrt{3} \approx 1.268 \), so the distance is approximately: \[ d = \frac{6}{1.268} \approx 4.73 \] which is \( \frac{15}{4} \). Thus, the correct answer is option (1), \( \frac{15}{4} \).