The length of the common chord of `x^2+y^2+2hx=0 , x^2+y^2-2ky=0` is

Options

A. `(hk)/sqrt(h^2-k^2)`
B. `(2hk)/sqrt(h^2+k^2)`
C. `(hk)/(h^2+k^2)`
D. `(2hk)/(h^2+k^2)`

Answer

To find the length of the common chord of the circles given by the equations \(x^2 + y^2 + 2hx = 0\) and \(x^2 + y^2 - 2ky = 0\), we can follow these steps:

### Step 1: Rewrite the equations of the circles
The first circle's equation can be rewritten as:
\[ x^2 + 2hx + y^2 = 0 \implies (x + h)^2 + y^2 = h^2 \]
This shows that the center of the first circle is \((-h, 0)\) and its radius is \(h\).

The second circle's equation can be rewritten as:
\[ x^2 + y^2 - 2ky = 0 \implies x^2 + (y - k)^2 = k^2 \]
This shows that the center of the second circle is \((0, k)\) and its radius is \(k\).

### Step 2: Find the equation of the common chord
The common chord can be found by subtracting the equations of the two circles:
\[ (x^2 + y^2 + 2hx) - (x^2 + y^2 - 2ky) = 0 \]
This simplifies to:
\[ 2hx + 2ky = 0 \implies hx + ky = 0 \]
This is the equation of the common chord.

### Step 3: Calculate the distance from the center of the first circle to the common chord
The distance \(d\) from the center of the first circle \((-h, 0)\) to the line \(hx + ky = 0\) can be calculated using the formula for the distance from a point to a line:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = h\), \(B = k\), \(C = 0\), and the point is \((-h, 0)\):
\[ d = \frac{|h(-h) + k(0) + 0|}{\sqrt{h^2 + k^2}} = \frac{h^2}{\sqrt{h^2 + k^2}} \]

### Step 4: Use Pythagoras' theorem to find the length of the common chord
Let \(AP\) be half the length of the common chord. By Pythagoras' theorem:
\[ O_1A^2 = OP^2 + AP^2 \]
Where \(O_1A\) is the radius of the first circle, which is \(h\), and \(OP\) is the distance we calculated:
\[ AP^2 = O_1A^2 - OP^2 = h^2 - \left(\frac{h^2}{\sqrt{h^2 + k^2}}\right)^2 \]
Calculating \(OP^2\):
\[ OP^2 = \frac{h^4}{h^2 + k^2} \]
Thus,
\[ AP^2 = h^2 - \frac{h^4}{h^2 + k^2} = \frac{h^2(h^2 + k^2) - h^4}{h^2 + k^2} = \frac{h^2k^2}{h^2 + k^2} \]
So,
\[ AP = \sqrt{\frac{h^2k^2}{h^2 + k^2}} = \frac{hk}{\sqrt{h^2 + k^2}} \]
The length of the common chord \(AB\) is \(2AP\):
\[ AB = 2AP = 2 \cdot \frac{hk}{\sqrt{h^2 + k^2}} = \frac{2hk}{\sqrt{h^2 + k^2}} \]

### Final Answer
The length of the common chord is:
\[ \frac{2hk}{\sqrt{h^2 + k^2}} \]