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 find the coordinates of the point of contact between the circle with center at (-1, 1) and the line given by the equation \(x + 2y = 12\), we can follow these steps:

### Step 1: Identify the center and the equation of the line
The center of the circle is given as \(C(-1, 1)\). The equation of the line can be rewritten in standard form as:
\[ x + 2y - 12 = 0 \]
Here, \(a = 1\), \(b = 2\), and \(c = -12\).

### Step 2: Calculate the distance from the center of the circle to the line
The distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by the formula:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Substituting the values:
- \(x_1 = -1\)
- \(y_1 = 1\)
- \(A = 1\)
- \(B = 2\)
- \(C = -12\)

We can calculate:
\[ d = \frac{|1(-1) + 2(1) - 12|}{\sqrt{1^2 + 2^2}} = \frac{|-1 + 2 - 12|}{\sqrt{1 + 4}} = \frac{|-11|}{\sqrt{5}} = \frac{11}{\sqrt{5}} \]

### Step 3: Find the foot of the perpendicular from the center to the line
The coordinates of the foot of the perpendicular can be found using the formula:
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{-Ax_1 - By_1 - C}{A^2 + B^2} \]
Substituting the known values:
- \(x_1 = -1\)
- \(y_1 = 1\)
- \(a = 1\)
- \(b = 2\)

We have:
\[ \frac{x + 1}{1} = \frac{y - 1}{2} = \frac{-1(-1) + 2(1) + 12}{1^2 + 2^2} \]
Calculating the right side:
\[ = \frac{1 + 2 + 12}{5} = \frac{15}{5} = 3 \]

Thus, we can set up the equations:
1. \(x + 1 = 3 \Rightarrow x = 2\)
2. \(y - 1 = 6 \Rightarrow y = 7\)

### Step 4: Coordinates of the point of contact
Thus, the coordinates of the point of contact \(P\) are:
\[ P(2, 7) \]

### Final Result
The coordinates of the point of contact are \((2, 7)\).

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