Find the equation of the tangent to the circle `x ^(2) + y^(2) =25` at the point in the first quadrnat where the diameter `4x- 3y =0` meets the circle.

Answer

To find the equation of the tangent to the circle \( x^2 + y^2 = 25 \) at the point in the first quadrant where the diameter \( 4x - 3y = 0 \) meets the circle, we can follow these steps:

### Step 1: Identify the Circle's Center and Radius
The equation of the circle is given by:
\[ x^2 + y^2 = 25 \]
From this, we can identify that the center of the circle is at the origin \( (0, 0) \) and the radius \( r \) is:
\[ r = \sqrt{25} = 5 \]

### Step 2: Find the Intersection Point of the Diameter and the Circle
The diameter is given by the equation:
\[ 4x - 3y = 0 \quad \Rightarrow \quad y = \frac{4}{3}x \]
We will substitute \( y \) in the circle's equation to find the intersection points.

Substituting \( y = \frac{4}{3}x \) into the circle's equation:
\[ x^2 + \left(\frac{4}{3}x\right)^2 = 25 \]
This simplifies to:
\[ x^2 + \frac{16}{9}x^2 = 25 \]
Combining the terms gives:
\[ \frac{25}{9}x^2 = 25 \]
Multiplying both sides by 9:
\[ 25x^2 = 225 \quad \Rightarrow \quad x^2 = 9 \quad \Rightarrow \quad x = 3 \quad (\text{since we are in the first quadrant}) \]
Now, substituting \( x = 3 \) back to find \( y \):
\[ y = \frac{4}{3} \cdot 3 = 4 \]
Thus, the point of intersection is \( (3, 4) \).

### Step 3: Find the Slope of the Tangent
To find the slope of the tangent at the point \( (3, 4) \), we need to differentiate the circle's equation:
\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25) \]
This gives:
\[ 2x + 2y \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{y} \]
Substituting \( x = 3 \) and \( y = 4 \):
\[ \frac{dy}{dx} = -\frac{3}{4} \]

### Step 4: Write the Equation of the Tangent Line
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (3, 4) \) and \( m = -\frac{3}{4} \):
\[ y - 4 = -\frac{3}{4}(x - 3) \]
Multiplying through by 4 to eliminate the fraction:
\[ 4(y - 4) = -3(x - 3) \]
Expanding both sides:
\[ 4y - 16 = -3x + 9 \]
Rearranging gives:
\[ 3x + 4y = 25 \]

### Final Equation of the Tangent
Thus, the equation of the tangent to the circle at the point \( (3, 4) \) is:
\[ \boxed{3x + 4y = 25} \]