(a,0) and (b, 0) are centres of two circles belonging to a co-axial system of which y-axis is the radical axis. If radius of one of the circles 'r', then the radius of the other circle is

Options

A. `(r^2+b^2+a^2)^(1//2)`
B. `(r^2+b^2-a^2)^(1//2)`
C. `(r^2+b^2-a^2)^(1//3)`
D. `(r^2+b^2+a^2)^(1//3)`

Answer

To solve the problem step by step, we need to find the radius \( r' \) of the second circle given that the first circle has a radius \( r \) and both circles have centers at points \( (a, 0) \) and \( (b, 0) \) respectively, with the y-axis as the radical axis.

### Step 1: Write the equations of the circles
The equation of the first circle with center \( (a, 0) \) and radius \( r \) is:
\[ (x - a)^2 + y^2 = r^2 \]
The equation of the second circle with center \( (b, 0) \) and radius \( r' \) is:
\[ (x - b)^2 + y^2 = r'^2 \]

### Step 2: Find the equation of the radical axis
The radical axis can be found by subtracting the equations of the two circles:
\[ (x - a)^2 + y^2 - [(x - b)^2 + y^2] = r^2 - r'^2 \]
This simplifies to:
\[ (x - a)^2 - (x - b)^2 = r^2 - r'^2 \]

### Step 3: Expand the left-hand side
Expanding the left-hand side:
\[ (x^2 - 2ax + a^2) - (x^2 - 2bx + b^2) = r^2 - r'^2 \]
This simplifies to:
\[ -2ax + a^2 + 2bx - b^2 = r^2 - r'^2 \]
Rearranging gives:
\[ (2b - 2a)x + (a^2 - b^2) = r^2 - r'^2 \]

### Step 4: Set the radical axis equal to the y-axis
Since the y-axis is the radical axis, we have:
\[ x = 0 \]
Substituting \( x = 0 \) into the equation gives:
\[ 0 + (a^2 - b^2) = r^2 - r'^2 \]
This simplifies to:
\[ a^2 - b^2 = r^2 - r'^2 \]

### Step 5: Rearranging to find \( r' \)
Rearranging the equation gives:
\[ r'^2 = r^2 + b^2 - a^2 \]
Taking the square root, we find:
\[ r' = \sqrt{r^2 + b^2 - a^2} \]

### Final Result
Thus, the radius of the other circle is:
\[ r' = \sqrt{r^2 + b^2 - a^2} \]