Consider the relation `4l^(2)-5m^(2)+6l+1=0` , where `l,m in R`
The number of tangents which can be drawn from the point (2,-3) to the above fixed circle are

Options

A. 0
B. 1
C. 2
D. 1 or 2

Explanation

To solve the problem of finding the number of tangents that can be drawn from the point (2, -3) to the circle defined by the equation \(4l^2 - 5m^2 + 6l + 1 = 0\), we will follow these steps:

### Step 1: Identify the Equation of the Circle
The given relation can be rewritten in the standard form of a circle. The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation \(4l^2 - 5m^2 + 6l + 1 = 0\), we can identify the coefficients:
- Coefficient of \(l^2\) (which is \(x^2\)) is \(4\).
- Coefficient of \(m^2\) (which is \(y^2\)) is \(-5\).
- Coefficient of \(l\) (which is \(2gx\)) is \(6\).
- Constant term \(c\) is \(1\).

### Step 2: Rewrite the Circle Equation
We can express the circle in the standard form:
\[ 4l^2 - 5m^2 + 6l + 1 = 0 \implies 4(l^2 + \frac{3}{2}l) - 5m^2 + 1 = 0 \]
Completing the square for \(l\):
\[ 4\left(l + \frac{3}{4}\right)^2 - 4\left(\frac{9}{16}\right) - 5m^2 + 1 = 0 \]
This simplifies to:
\[ 4\left(l + \frac{3}{4}\right)^2 - 5m^2 - \frac{1}{4} = 0 \]
Now, we can express it as:
\[ 4\left(l + \frac{3}{4}\right)^2 + 5m^2 = \frac{1}{4} \]

### Step 3: Identify the Center and Radius
From the equation \(4\left(l + \frac{3}{4}\right)^2 + 5m^2 = \frac{1}{4}\), we can identify:
- The center of the circle is \((-g, -f) = \left(-\frac{3}{4}, 0\right)\).
- The radius \(r\) can be found from the equation:
\[ r = \sqrt{g^2 + f^2 - c} \]
In this case, we need to find the radius from the standard form, which is not straightforward from the above equation. However, we can find the radius from the transformed equation.

### Step 4: Calculate the Distance from the Point to the Center
The distance \(d\) from the point \(P(2, -3)\) to the center of the circle \(\left(-\frac{3}{4}, 0\right)\) is given by:
\[ d = \sqrt{\left(2 + \frac{3}{4}\right)^2 + (-3 - 0)^2} \]
Calculating this:
\[ d = \sqrt{\left(\frac{8}{4} + \frac{3}{4}\right)^2 + (-3)^2} = \sqrt{\left(\frac{11}{4}\right)^2 + 9} = \sqrt{\frac{121}{16} + \frac{144}{16}} = \sqrt{\frac{265}{16}} = \frac{\sqrt{265}}{4} \]

### Step 5: Compare the Distance with the Radius
Now we compare \(d\) with the radius \(r\). If \(d > r\), there are 2 tangents; if \(d = r\), there is 1 tangent; and if \(d < r\), there are no tangents.

### Step 6: Conclusion
Since we found that \(d > r\) (as the radius is very small compared to the distance), we conclude that there are 2 tangents that can be drawn from the point (2, -3) to the circle.

### Final Answer
The number of tangents that can be drawn from the point (2, -3) to the circle is **2**.

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