If the line `x+2y=3` cuts a chord of length r unit with the circle `x^2+y^2=r^2` then find `r^2`.

Options

A. `12/5`
B. `sqrt(12)`
C. `5/12`
D. `sqrt(12/5)`

Answer

To solve the problem, we need to find \( r^2 \) where the line \( x + 2y = 3 \) cuts a chord of length \( r \) with the circle \( x^2 + y^2 = r^2 \).

### Step 1: Identify the Circle and Line
The equation of the circle is given by:
\[ x^2 + y^2 = r^2 \]
This circle has its center at the origin \( (0, 0) \) and a radius of \( r \).

The equation of the line can be rewritten as:
\[ x + 2y - 3 = 0 \]

### Step 2: Find the Distance from the Center of the Circle to the Line
The distance \( d \) from a point \( (x_1, y_1) \) to a line \( ax + by + c = 0 \) is given by the formula:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]

For our line \( x + 2y - 3 = 0 \):
- \( a = 1 \)
- \( b = 2 \)
- \( c = -3 \)

The center of the circle is \( (0, 0) \), so we substitute \( x_1 = 0 \) and \( y_1 = 0 \):
\[ d = \frac{|1(0) + 2(0) - 3|}{\sqrt{1^2 + 2^2}} = \frac{|-3|}{\sqrt{1 + 4}} = \frac{3}{\sqrt{5}} \]

### Step 3: Relate the Distance to the Chord Length
The distance \( d \) from the center to the line is related to the radius \( r \) and half the length of the chord \( \frac{L}{2} \) using the Pythagorean theorem:
\[ r^2 = d^2 + \left(\frac{L}{2}\right)^2 \]
Here, \( L = r \) (the length of the chord), so \( \frac{L}{2} = \frac{r}{2} \).

Substituting \( d \) into the equation:
\[ r^2 = \left(\frac{3}{\sqrt{5}}\right)^2 + \left(\frac{r}{2}\right)^2 \]

### Step 4: Simplify the Equation
Calculating \( d^2 \):
\[ \left(\frac{3}{\sqrt{5}}\right)^2 = \frac{9}{5} \]
Now substituting back into the equation:
\[ r^2 = \frac{9}{5} + \frac{r^2}{4} \]

### Step 5: Solve for \( r^2 \)
To eliminate the fraction, multiply through by 20 (the least common multiple of 5 and 4):
\[ 20r^2 = 36 + 5r^2 \]
Rearranging gives:
\[ 20r^2 - 5r^2 = 36 \]
\[ 15r^2 = 36 \]
\[ r^2 = \frac{36}{15} = \frac{12}{5} \]

### Final Answer
Thus, the value of \( r^2 \) is:
\[ \boxed{\frac{12}{5}} \]