If x+2y=3 and 2x+y=3 touches a circle whose centres lies on 3x+4y=5, then possible co-ordinates of centres of circle are

Options

A. `(4/3,1/4)`
B. `(-3,2)`
C. `(5/7,5/7)`
D. `(3,-1)`

Answer

To solve the problem step-by-step, we will follow the reasoning laid out in the video transcript.

### Step 1: Set Up the Problem
We have two lines given by the equations:
1. \( x + 2y = 3 \) (Equation 1)
2. \( 2x + y = 3 \) (Equation 2)

We also know that the center of the circle lies on the line given by:
3. \( 3x + 4y = 5 \) (Equation 3)

Let the coordinates of the center of the circle be \( (h, k) \).

### Step 2: Substitute the Center into the Line Equation
Since the center lies on the line \( 3x + 4y = 5 \), we substitute \( h \) and \( k \) into this equation:
\[ 3h + 4k = 5 \quad \text{(Equation 4)} \]

### Step 3: Find the Perpendicular Distances
The distance from the center \( (h, k) \) to the line \( x + 2y - 3 = 0 \) (which can be rewritten from Equation 1) is given by the formula:
\[ \text{Distance} = \frac{|h + 2k - 3|}{\sqrt{1^2 + 2^2}} = \frac{|h + 2k - 3|}{\sqrt{5}} \]

The distance from the center \( (h, k) \) to the line \( 2x + y - 3 = 0 \) (from Equation 2) is:
\[ \text{Distance} = \frac{|2h + k - 3|}{\sqrt{2^2 + 1^2}} = \frac{|2h + k - 3|}{\sqrt{5}} \]

### Step 4: Set the Distances Equal
Since both lines are tangents to the circle, the distances must be equal:
\[ |h + 2k - 3| = |2h + k - 3| \quad \text{(Equation 5)} \]

### Step 5: Solve the Absolute Value Equation
This gives us two cases to solve:

**Case 1:**
\[ h + 2k - 3 = 2h + k - 3 \]
Simplifying this:
\[ h + 2k = 2h + k \implies k = h \]

**Case 2:**
\[ h + 2k - 3 = -(2h + k - 3) \]
Simplifying this:
\[ h + 2k - 3 = -2h - k + 3 \]
\[ h + 2k + 2h + k = 6 \implies 3h + 3k = 6 \implies h + k = 2 \quad \text{(Equation 6)} \]

### Step 6: Substitute Back into Equation 4
Now we have two equations:
1. \( k = h \) (from Case 1)
2. \( h + k = 2 \) (from Case 2)

Substituting \( k = h \) into Equation 4:
\[ 3h + 4h = 5 \implies 7h = 5 \implies h = \frac{5}{7}, \quad k = \frac{5}{7} \]
Thus, one possible center is \( \left( \frac{5}{7}, \frac{5}{7} \right) \).

### Step 7: Solve for the Second Center
Now substituting \( h + k = 2 \) into Equation 4:
Let \( k = 2 - h \):
\[ 3h + 4(2 - h) = 5 \]
\[ 3h + 8 - 4h = 5 \implies -h + 8 = 5 \implies -h = -3 \implies h = 3 \]
Then substituting back to find \( k \):
\[ k = 2 - 3 = -1 \]
Thus, the second possible center is \( (3, -1) \).

### Final Answer
The possible coordinates of the centers of the circle are:
1. \( \left( \frac{5}{7}, \frac{5}{7} \right) \)
2. \( (3, -1) \)