A tangent is drawn to the circle `2(x^(2)+y^(2))-3x+4y=0` and it touches the circle at point A. If the tangent passes through the point P(2, 1),then PA=

Options

A. 4
B. 2
C. `2sqrt(2)`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Rewrite the equation of the circle
The given equation of the circle is:
\[ 2(x^2 + y^2) - 3x + 4y = 0 \]
Dividing the entire equation by 2 gives:
\[ x^2 + y^2 - \frac{3}{2}x + 2y = 0 \]

### Step 2: Complete the square for the x and y terms
To rewrite the equation in standard form, we complete the square for both the x and y terms.

For \( x \):
\[ x^2 - \frac{3}{2}x \]
To complete the square:
\[ = \left(x - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 = \left(x - \frac{3}{4}\right)^2 - \frac{9}{16} \]

For \( y \):
\[ y^2 + 2y \]
To complete the square:
\[ = (y + 1)^2 - 1 \]

Putting it all together, we have:
\[ \left(x - \frac{3}{4}\right)^2 - \frac{9}{16} + (y + 1)^2 - 1 = 0 \]
This simplifies to:
\[ \left(x - \frac{3}{4}\right)^2 + (y + 1)^2 = \frac{25}{16} \]

### Step 3: Identify the center and radius of the circle
From the standard form of the circle:
- Center \( C \left(\frac{3}{4}, -1\right) \)
- Radius \( R = \sqrt{\frac{25}{16}} = \frac{5}{4} \)

### Step 4: Find the distance from point P to the center of the circle
The point \( P \) is given as \( (2, 1) \). We can find the distance \( d \) from point \( P \) to the center \( C \):
\[ d = \sqrt{\left(2 - \frac{3}{4}\right)^2 + \left(1 - (-1)\right)^2} \]
Calculating the x-component:
\[ 2 - \frac{3}{4} = \frac{8}{4} - \frac{3}{4} = \frac{5}{4} \]
Calculating the y-component:
\[ 1 - (-1) = 1 + 1 = 2 \]
Now substituting these values back into the distance formula:
\[ d = \sqrt{\left(\frac{5}{4}\right)^2 + 2^2} = \sqrt{\frac{25}{16} + 4} = \sqrt{\frac{25}{16} + \frac{64}{16}} = \sqrt{\frac{89}{16}} = \frac{\sqrt{89}}{4} \]

### Step 5: Use the Pythagorean theorem to find PA
Using the Pythagorean theorem, we know:
\[ PA^2 + R^2 = d^2 \]
Where \( PA \) is the distance from point \( P \) to point \( A \) (the point of tangency). We can rearrange this to find \( PA \):
\[ PA^2 = d^2 - R^2 \]
Substituting the values we found:
\[ d^2 = \frac{89}{16}, \quad R^2 = \left(\frac{5}{4}\right)^2 = \frac{25}{16} \]
Thus:
\[ PA^2 = \frac{89}{16} - \frac{25}{16} = \frac{64}{16} = 4 \]
Taking the square root gives:
\[ PA = 2 \]

### Final Answer
The distance \( PA \) is \( 2 \).

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