To find the nearest point on the circle defined by the equation \(x^2 + y^2 - 6x + 4y - 12 = 0\) from the point \((-5, 4)\), we will follow these steps:
### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. We can do this by completing the square for both \(x\) and \(y\).
The given equation is:
\[
x^2 + y^2 - 6x + 4y - 12 = 0
\]
Rearranging gives:
\[
x^2 - 6x + y^2 + 4y = 12
\]
Now, we complete the square:
1. For \(x^2 - 6x\):
- Take half of \(-6\) (which is \(-3\)), square it to get \(9\).
- Thus, \(x^2 - 6x = (x - 3)^2 - 9\).
2. For \(y^2 + 4y\):
- Take half of \(4\) (which is \(2\)), square it to get \(4\).
- Thus, \(y^2 + 4y = (y + 2)^2 - 4\).
Substituting back into the equation gives:
\[
(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12
\]
\[
(x - 3)^2 + (y + 2)^2 - 13 = 12
\]
\[
(x - 3)^2 + (y + 2)^2 = 25
\]
### Step 2: Identify the Center and Radius
From the standard form \((x - 3)^2 + (y + 2)^2 = 25\), we can identify:
- Center of the circle \(C(3, -2)\)
- Radius \(r = \sqrt{25} = 5\)
### Step 3: Find the Distance from the Point to the Center
Now we need to find the distance from the point \((-5, 4)\) to the center \(C(3, -2)\).
Using the distance formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
where \((x_1, y_1) = (-5, 4)\) and \((x_2, y_2) = (3, -2)\):
\[
d = \sqrt{(3 - (-5))^2 + (-2 - 4)^2}
\]
\[
= \sqrt{(3 + 5)^2 + (-2 - 4)^2}
\]
\[
= \sqrt{8^2 + (-6)^2}
\]
\[
= \sqrt{64 + 36} = \sqrt{100} = 10
\]
### Step 4: Determine the Nearest Point on the Circle
The distance from the center \(C\) to the nearest point \(R\) on the circle is equal to the radius \(r = 5\).
Since the total distance from the point \((-5, 4)\) to the center \(C(3, -2)\) is \(10\), we can find the distance from the center to the nearest point on the circle:
\[
PC = 10 \quad \text{and} \quad RC = 5
\]
Thus, the distance from the point to the nearest point on the circle is:
\[
PR = PC - RC = 10 - 5 = 5
\]
### Step 5: Find the Coordinates of the Nearest Point
To find the coordinates of the nearest point \(R\), we need to move from the center \(C(3, -2)\) towards the point \((-5, 4)\) by a distance of \(5\).
First, we find the direction vector from \(C\) to the point \((-5, 4)\):
\[
\text{Direction vector} = (-5 - 3, 4 - (-2)) = (-8, 6)
\]
Next, we find the unit vector in this direction:
\[
\text{Magnitude} = \sqrt{(-8)^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10
\]
\[
\text{Unit vector} = \left(-\frac{8}{10}, \frac{6}{10}\right) = \left(-0.8, 0.6\right)
\]
Now, we move from the center \(C(3, -2)\) in the direction of the unit vector scaled by \(5\):
\[
R = C + 5 \times \text{Unit vector} = (3, -2) + 5 \times (-0.8, 0.6)
\]
\[
= (3 - 4, -2 + 3) = (-1, 1)
\]
### Final Answer
The nearest point on the circle from the point \((-5, 4)\) is:
\[
\boxed{(-1, 1)}
\]
