Equation of circle centred at origin , and touching the line
` 3x - 4y + 20 =0 ` is

Options

A. `x^(2) + y^(2) = 4 `
B. `x^(2) + y^(2) = 9 `
C. `x^(2) + y^(2) = 25 `
D. `x^(2) + y^(2) = 16`

Answer

To find the equation of a circle centered at the origin (0, 0) that touches the line given by the equation \(3x - 4y + 20 = 0\), we can follow these steps:

### Step 1: Identify the line's coefficients
The equation of the line can be written in the standard form \(Ax + By + C = 0\). Here, we have:
- \(A = 3\)
- \(B = -4\)
- \(C = 20\)

### Step 2: Calculate the perpendicular distance from the center of the circle to the line
The formula for the perpendicular distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Since the center of the circle is at the origin \((0, 0)\), we substitute \(x_1 = 0\) and \(y_1 = 0\):
\[ d = \frac{|3(0) + (-4)(0) + 20|}{\sqrt{3^2 + (-4)^2}} = \frac{|20|}{\sqrt{9 + 16}} = \frac{20}{\sqrt{25}} = \frac{20}{5} = 4 \]

### Step 3: Determine the radius of the circle
Since the circle touches the line, the radius \(r\) of the circle is equal to the perpendicular distance calculated in Step 2. Therefore, the radius \(r = 4\).

### Step 4: Write the equation of the circle
The standard equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
For our circle centered at the origin \((0, 0)\) and with radius \(4\), the equation becomes:
\[ (x - 0)^2 + (y - 0)^2 = 4^2 \]
This simplifies to:
\[ x^2 + y^2 = 16 \]

### Final Answer
The equation of the circle is:
\[ \boxed{x^2 + y^2 = 16} \]

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