Find the equation of a circle passes through the origin and whose centre is (-2,5).

Answer

To find the equation of the circle that passes through the origin and has its center at (-2, 5), we can follow these steps:

### Step 1: Write the standard form of the circle's equation
The standard form of the equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Step 2: Substitute the center coordinates
Given that the center of the circle is \((-2, 5)\), we substitute \(h = -2\) and \(k = 5\) into the equation:
\[ (x + 2)^2 + (y - 5)^2 = r^2 \]

### Step 3: Use the point that the circle passes through
Since the circle passes through the origin \((0, 0)\), we can substitute \(x = 0\) and \(y = 0\) into the equation to find \(r^2\):
\[ (0 + 2)^2 + (0 - 5)^2 = r^2 \]

### Step 4: Calculate \(r^2\)
Now, we simplify the left-hand side:
\[ 2^2 + (-5)^2 = r^2 \]
\[ 4 + 25 = r^2 \]
\[ r^2 = 29 \]

### Step 5: Write the complete equation of the circle
Now that we have \(r^2\), we can substitute it back into the equation of the circle:
\[ (x + 2)^2 + (y - 5)^2 = 29 \]

### Step 6: Expand the equation
To express the equation in a standard polynomial form, we expand it:
\[ (x + 2)^2 = x^2 + 4x + 4 \]
\[ (y - 5)^2 = y^2 - 10y + 25 \]
Combining these gives:
\[ x^2 + 4x + 4 + y^2 - 10y + 25 = 29 \]

### Step 7: Simplify the equation
Now, we simplify the equation:
\[ x^2 + y^2 + 4x - 10y + 29 = 29 \]
Subtracting 29 from both sides results in:
\[ x^2 + y^2 + 4x - 10y = 0 \]

### Final Answer
Thus, the equation of the circle is:
\[ x^2 + y^2 + 4x - 10y = 0 \]

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