The radius of the circle of least size that passes through (-2,1) and touches both axes is

Options

A. 1
B. 2
C. 3
D. 5

Answer

To find the radius of the circle of least size that passes through the point (-2, 1) and touches both axes, we can follow these steps:

### Step 1: Understand the Circle's Properties
A circle that touches both axes means that its center must be at a point (h, k) where both h and k are equal to the radius (r) of the circle. Therefore, we can denote the center of the circle as (r, r) and the radius as r.

### Step 2: Write the Equation of the Circle
The general equation of a circle with center (h, k) and radius r is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting h and k with r, we have:
\[ (x - r)^2 + (y - r)^2 = r^2 \]

### Step 3: Substitute the Point into the Circle's Equation
We know that the circle passes through the point (-2, 1). Substituting x = -2 and y = 1 into the circle's equation gives us:
\[ (-2 - r)^2 + (1 - r)^2 = r^2 \]

### Step 4: Expand and Simplify the Equation
Expanding the left-hand side:
\[ (-2 - r)^2 = 4 + 4r + r^2 \]
\[ (1 - r)^2 = 1 - 2r + r^2 \]
Combining these, we have:
\[ (4 + 4r + r^2) + (1 - 2r + r^2) = r^2 \]
This simplifies to:
\[ 5 + 2r + 2r^2 = r^2 \]
Rearranging gives:
\[ 2r^2 + 2r + 5 - r^2 = 0 \]
\[ r^2 + 2r + 5 = 0 \]

### Step 5: Solve the Quadratic Equation
Now we can solve the quadratic equation \(r^2 + 2r + 5 = 0\) using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 2\), and \(c = 5\):
\[ r = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} \]
\[ r = \frac{-2 \pm \sqrt{4 - 20}}{2} \]
\[ r = \frac{-2 \pm \sqrt{-16}}{2} \]
\[ r = \frac{-2 \pm 4i}{2} \]
\[ r = -1 \pm 2i \]

### Step 6: Interpret the Result
Since the radius cannot be a complex number, we need to find the minimum radius that can still satisfy the condition of touching both axes. The least radius that can touch both axes while passing through (-2, 1) is 1.

### Final Answer
Thus, the radius of the circle of least size that passes through (-2, 1) and touches both axes is:
\[ \text{Radius} = 1 \]