The shortest distance of (-5,4) to the circle `x^(2)+y^(2)-6x+4y-12=0` is

Options

A. 10
B. 3
C. 5
D. 4

Answer

To find the shortest distance from the point (-5, 4) to the circle given by the equation \(x^2 + y^2 - 6x + 4y - 12 = 0\), we will follow these steps:

### Step 1: Rewrite the Circle's Equation
We start with the equation of the circle:
\[ x^2 + y^2 - 6x + 4y - 12 = 0 \]
We can rearrange this into the standard form of a circle by completing the square.

### Step 2: Completing the Square
1. For the \(x\) terms: \(x^2 - 6x\)
- Take half of -6, square it: \((-3)^2 = 9\)
- Rewrite: \(x^2 - 6x = (x - 3)^2 - 9\)

2. For the \(y\) terms: \(y^2 + 4y\)
- Take half of 4, square it: \((2)^2 = 4\)
- Rewrite: \(y^2 + 4y = (y + 2)^2 - 4\)

Now substitute these back into the equation:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0 \]
Combine the constants:
\[ (x - 3)^2 + (y + 2)^2 - 25 = 0 \]
Thus, the equation of the circle in standard form is:
\[ (x - 3)^2 + (y + 2)^2 = 25 \]

### Step 3: Identify the Center and Radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\):
- Center \(C = (3, -2)\)
- Radius \(r = \sqrt{25} = 5\)

### Step 4: Calculate the Distance from the Point to the Center
Now we need to find the distance from the point \((-5, 4)\) to the center of the circle \((3, -2)\):
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(3 - (-5))^2 + (-2 - 4)^2} \]
\[ = \sqrt{(3 + 5)^2 + (-2 - 4)^2} \]
\[ = \sqrt{8^2 + (-6)^2} \]
\[ = \sqrt{64 + 36} \]
\[ = \sqrt{100} \]
\[ = 10 \]

### Step 5: Calculate the Shortest Distance to the Circle
The shortest distance from the point to the circle is the distance from the point to the center minus the radius:
\[ \text{Shortest Distance} = d - r = 10 - 5 = 5 \]

### Final Answer
The shortest distance from the point (-5, 4) to the circle is **5**.

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