Question

Length of the common chord of the circles \(x^2 + y^2 - 6x + 5 = 0\) and \(x^2 + y^2 + 4y - 5 = 0\) is:

• A. \(\sqrt{13}\)
• B. \(\frac{12}{\sqrt{13}}\)
• C. \(\frac{6}{\sqrt{13}}\)
• D. \(2\sqrt{13}\)

Hint:

For the common chord of two circles, use the distance between their centers and the radii to calculate the chord length.
- Use the formula: \(L = 2 \sqrt{r_1^2 - \left(\frac{d^2 - r_2^2 + r_1^2}{2d}\right)^2}\).

Answer 1

The Correct Option is B

We are given two circles with equations: 1. \(x^2 + y^2 - 6x + 5 = 0\) 2. \(x^2 + y^2 + 4y - 5 = 0\) ### Step 1: Rewriting the equations in standard form. For the first circle, complete the square for \(x\): \[ x^2 - 6x + 9 + y^2 + 5 - 9 = 0 \quad \Rightarrow \quad (x - 3)^2 + y^2 = 4. \] So, the center of the first circle is \((3, 0)\) and the radius is \(2\). For the second circle, complete the square for \(y\): \[ x^2 + y^2 + 4y + 4 - 4 - 5 = 0 \quad \Rightarrow \quad x^2 + (y + 2)^2 = 1. \] So, the center of the second circle is \((0, -2)\) and the radius is \(1\). Step 2: Finding the length of the common chord. The distance \(d\) between the centers of the two circles is: \[ d = \sqrt{(3 - 0)^2 + (0 - (-2))^2} = \sqrt{9 + 4} = \sqrt{13}. \] Using the formula for the length \(L\) of the common chord: \[ L = \sqrt{r_1^2 - \left(\frac{d^2 - r_2^2 + r_1^2}{2d}\right)^2}, \] where \(r_1 = 2\), \(r_2 = 1\), and \(d = \sqrt{13}\). Substitute these values into the formula: \[ L = \sqrt{2^2 - \left(\frac{13 - 1 + 4}{2\sqrt{13}}\right)^2} = 2 \sqrt{4 - \left(\frac{16}{2\sqrt{13}}\right)^2} = 2 \sqrt{4 - \frac{256}{52}}. \] Simplifying this expression: \[ L = \sqrt{4 - \frac{64}{13}} = \sqrt{\frac{52}{13} - \frac{64}{13}} = \frac{12}{\sqrt{13}}. \] Thus, the length of the common chord is \(\boxed{\frac{12}{\sqrt{13}}}\).