If a circle, whose centre is (-1,1) touches the straight line x+2y = 12, then the co-ordinates of the point of contact are

Options

A. `(-(7)/(2),-4)`
B. `((6)/(5),(27)/(5))`
C. (2,-7)
D. (-2,-5)

Answer

To solve the problem, we need to find the coordinates of the point of contact where the circle touches the line. Here are the steps to derive the solution:

### Step 1: Identify the given information
The center of the circle is given as \( C(-1, 1) \) and the equation of the line is \( x + 2y = 12 \).

### Step 2: Write the equation of the line in slope-intercept form
We can rewrite the equation of the line to find its slope:
\[ x + 2y = 12 \implies 2y = -x + 12 \implies y = -\frac{1}{2}x + 6 \]
The slope of the line \( m_1 \) is \( -\frac{1}{2} \).

### Step 3: Find the slope of the normal to the circle
The slope of the radius (which is normal to the tangent at the point of contact) is the negative reciprocal of the slope of the tangent line. Therefore:
\[ m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{1}{2}} = 2 \]

### Step 4: Write the equation of the normal line
The equation of the normal line passing through the center \( C(-1, 1) \) with slope \( 2 \) can be written using the point-slope form:
\[ y - 1 = 2(x + 1) \]
Simplifying this gives:
\[ y - 1 = 2x + 2 \implies y = 2x + 3 \]

### Step 5: Find the point of contact
Since the point of contact \( P(a, b) \) lies on both the line \( x + 2y = 12 \) and the normal \( y = 2x + 3 \), we can substitute \( y \) from the normal line equation into the line equation:
\[ x + 2(2x + 3) = 12 \]
This simplifies to:
\[ x + 4x + 6 = 12 \implies 5x + 6 = 12 \implies 5x = 6 \implies x = \frac{6}{5} \]

### Step 6: Calculate the corresponding y-coordinate
Now, substituting \( x = \frac{6}{5} \) back into the normal line equation to find \( y \):
\[ y = 2\left(\frac{6}{5}\right) + 3 = \frac{12}{5} + 3 = \frac{12}{5} + \frac{15}{5} = \frac{27}{5} \]

### Step 7: State the coordinates of the point of contact
Thus, the coordinates of the point of contact are:
\[ P\left(\frac{6}{5}, \frac{27}{5}\right) \]

### Final Answer
The coordinates of the point of contact are \( \left(\frac{6}{5}, \frac{27}{5}\right) \).

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