The point of tangency of the circles `x^2+ y^2 - 2x-4y = 0 and x^2 + y^2-8y -4 = 0`, is

Options

A. touch each other intermally
B. touch each other externally
C. cuts each other at two points
D. None of these

Answer

To find the point of tangency of the circles given by the equations \(x^2 + y^2 - 2x - 4y = 0\) and \(x^2 + y^2 - 8y - 4 = 0\), we will follow these steps:

### Step 1: Rewrite the equations in standard form
We will convert both circle equations into the standard form \((x - h)^2 + (y - k)^2 = r^2\).

**For the first circle:**
\[ x^2 + y^2 - 2x - 4y = 0 \]
Rearranging gives:
\[ x^2 - 2x + y^2 - 4y = 0 \]
Completing the square:
\[ (x^2 - 2x + 1) + (y^2 - 4y + 4) = 5 \]
This simplifies to:
\[ (x - 1)^2 + (y - 2)^2 = 5 \]
So, the center \(C_1\) is \((1, 2)\) and the radius \(r_1 = \sqrt{5}\).

**For the second circle:**
\[ x^2 + y^2 - 8y - 4 = 0 \]
Rearranging gives:
\[ x^2 + (y^2 - 8y) = 4 \]
Completing the square:
\[ x^2 + (y^2 - 8y + 16) = 20 \]
This simplifies to:
\[ (x - 0)^2 + (y - 4)^2 = 20 \]
So, the center \(C_2\) is \((0, 4)\) and the radius \(r_2 = \sqrt{20} = 2\sqrt{5}\).

### Step 2: Calculate the distance between the centers
Using the distance formula, we calculate the distance \(C_1C_2\):
\[ C_1C_2 = \sqrt{(1 - 0)^2 + (2 - 4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Step 3: Check the condition for tangency
For two circles to touch each other internally, the distance between their centers \(C_1C_2\) must equal the absolute difference of their radii:
\[ C_1C_2 = |r_1 - r_2| \]
Substituting the values:
\[ \sqrt{5} = |\sqrt{5} - 2\sqrt{5}| \]
Calculating the right side:
\[ |\sqrt{5} - 2\sqrt{5}| = |-\sqrt{5}| = \sqrt{5} \]
Since both sides are equal, the circles touch each other internally.

### Step 4: Find the point of tangency
To find the point of tangency, we can use the section formula. The point of tangency divides the line segment joining the centers \(C_1\) and \(C_2\) in the ratio of their radii.

The ratio of the radii is:
\[ \frac{r_1}{r_2} = \frac{\sqrt{5}}{2\sqrt{5}} = \frac{1}{2} \]

Using the section formula, the coordinates of the point of tangency \(P\) dividing \(C_1C_2\) in the ratio \(1:2\) are given by:
\[ P = \left(\frac{1 \cdot 0 + 2 \cdot 1}{1 + 2}, \frac{1 \cdot 4 + 2 \cdot 2}{1 + 2}\right) = \left(\frac{2}{3}, \frac{4 + 4}{3}\right) = \left(\frac{2}{3}, \frac{8}{3}\right) \]

### Final Answer:
The point of tangency of the circles is \(\left(\frac{2}{3}, \frac{8}{3}\right)\).

---