O is the origin and A is a fixed point on the circle of radius 'a' with centre O.The vector `vec O A` is denoted by `vec a` . A variable point P lie on the tangent at `A and vec OP-vec r.` Show that `vec a vec r=a^2.`Hence if `P(x, y) and A(x_1, y_1),` deduce the equation of tangent at A to this circle.

Answer

To solve the given problem, we will follow these steps:

### Step 1: Understand the given information
We have:
- O as the origin (0, 0).
- A as a fixed point on the circle of radius 'a' with center O, denoted by the vector \(\vec{OA} = \vec{a}\).
- A variable point P lies on the tangent at A, and we denote \(\vec{OP} = \vec{r}\).

### Step 2: Establish the relationship between the vectors
Since point P lies on the tangent at point A, we can express the relationship between the vectors as follows:
\[ \vec{OP} - \vec{OA} = \vec{AP} \]
This means:
\[ \vec{AP} = \vec{r} - \vec{a} \]

### Step 3: Use the property of perpendicular vectors
Since the tangent at point A is perpendicular to the radius OA, we have:
\[ \vec{AP} \cdot \vec{OA} = 0 \]
Substituting \(\vec{AP}\) from the previous step, we get:
\[ (\vec{r} - \vec{a}) \cdot \vec{a} = 0 \]
Expanding this gives:
\[ \vec{r} \cdot \vec{a} - \vec{a} \cdot \vec{a} = 0 \]

### Step 4: Simplify the equation
Recall that \(\vec{a} \cdot \vec{a} = |\vec{a}|^2 = a^2\) since \(\vec{a}\) is the radius of the circle. Therefore, we can rewrite the equation as:
\[ \vec{r} \cdot \vec{a} = a^2 \]

### Step 5: Conclusion for the first part
Thus, we have shown that:
\[ \vec{a} \cdot \vec{r} = a^2 \]

### Step 6: Find the equation of the tangent at A
Let the coordinates of point A be \((x_1, y_1)\) and the coordinates of point P be \((x, y)\). The equation of the tangent line at point A can be derived from the dot product we established:
\[ \vec{OA} \cdot \vec{OP} = a^2 \]
In terms of coordinates, this translates to:
\[ x_1 x + y_1 y = a^2 \]

### Final Result
The equation of the tangent at point A to the circle is:
\[ x_1 x + y_1 y = a^2 \]

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