Equation of circle passing through the centre of the circle `x^2+y^2-4x-6y-8=0` and being concentric with the circle `x^2+y^2-2x-8y-5=0` is

Answer

To find the equation of a circle that passes through the center of the circle given by the equation \(x^2 + y^2 - 4x - 6y - 8 = 0\) and is concentric with the circle given by the equation \(x^2 + y^2 - 2x - 8y - 5 = 0\), we can follow these steps:

### Step 1: Identify the center of the first circle
The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the first circle's equation \(x^2 + y^2 - 4x - 6y - 8 = 0\), we can identify:
- \(g = -2\)
- \(f = -3\)

The center of the circle is given by \((-g, -f)\):
\[ \text{Center}_1 = (2, 3) \]

### Step 2: Identify the center of the second circle
From the second circle's equation \(x^2 + y^2 - 2x - 8y - 5 = 0\), we have:
- \(g = -1\)
- \(f = -4\)

The center of the second circle is:
\[ \text{Center}_2 = (2, 4) \]

### Step 3: Find the radius of the required circle
Since the required circle is concentric with the second circle, it will have the same center as the second circle, which is \((2, 4)\). The required circle must also pass through the center of the first circle \((2, 3)\).

We can find the radius \(r\) by calculating the distance between the two centers:
\[ r = \sqrt{(2 - 2)^2 + (4 - 3)^2} = \sqrt{0 + 1} = \sqrt{1} = 1 \]

### Step 4: Write the equation of the required circle
The equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 2\), \(k = 4\), and \(r = 1\):
\[ (x - 2)^2 + (y - 4)^2 = 1^2 \]
Thus, the equation of the required circle is:
\[ (x - 2)^2 + (y - 4)^2 = 1 \]

### Final Answer
The equation of the circle is:
\[ (x - 2)^2 + (y - 4)^2 = 1 \]

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