`(x-2)^2+(y+5)^2=36`
IF a circle in the xy-plane has the equation above. Which of the following does NOT lie on the exterior of the circle?

Options

A. (2,1)
B. (2,5)
C. (5,2)
D. (-1,1)

Answer

To determine which point does NOT lie on the exterior of the circle given by the equation \((x-2)^2+(y+5)^2=36\), we can follow these steps:

### Step 1: Identify the center and radius of the circle
The general form of the equation of a circle is \((x - a)^2 + (y - b)^2 = r^2\), where \((a, b)\) is the center and \(r\) is the radius.

From the equation \((x - 2)^2 + (y + 5)^2 = 36\):
- The center \((a, b)\) is \((2, -5)\).
- The radius \(r\) is \(\sqrt{36} = 6\).

### Step 2: Use the distance formula to check points
To check whether a point \((x_1, y_1)\) lies inside, on, or outside the circle, we can use the following condition:
- Calculate \(d = (x_1 - 2)^2 + (y_1 + 5)^2\).
- Compare \(d\) with \(r^2\) (which is 36).

- If \(d < 36\), the point is inside the circle.
- If \(d = 36\), the point is on the circle.
- If \(d > 36\), the point is outside the circle.

### Step 3: Check each point
Let's check the given points one by one:

1. **Point (2, 1)**:
\[ d = (2 - 2)^2 + (1 + 5)^2 = 0 + 6^2 = 36 \]
Since \(d = 36\), this point lies on the circle.

2. **Point (2, 5)**:
\[ d = (2 - 2)^2 + (5 + 5)^2 = 0 + 10^2 = 100 \]
Since \(d > 36\), this point lies outside the circle.

3. **Point (5, 2)**:
\[ d = (5 - 2)^2 + (2 + 5)^2 = 3^2 + 7^2 = 9 + 49 = 58 \]
Since \(d > 36\), this point lies outside the circle.

4. **Point (-1, 1)**:
\[ d = (-1 - 2)^2 + (1 + 5)^2 = (-3)^2 + 6^2 = 9 + 36 = 45 \]
Since \(d > 36\), this point lies outside the circle.

### Conclusion
The only point that does NOT lie on the exterior of the circle is **(2, 1)**, as it lies on the circle.

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