The two circles `x^(2)+y^(2)-2x-2y-7=0` and `3(x^(2)+y^(2))-8x+29y=0`

Options

A. touch externally
B. touch internally
C. cut each other orthogonally
D. do not cut each other

Answer

To determine the relationship between the two circles given by the equations \(x^2 + y^2 - 2x - 2y - 7 = 0\) and \(3(x^2 + y^2) - 8x + 29y = 0\), we will follow these steps:

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

**For the first circle:**
Starting with the equation:
\[ x^2 + y^2 - 2x - 2y - 7 = 0 \]
We can rearrange it:
\[ x^2 - 2x + y^2 - 2y = 7 \]
Now, we complete the square for \(x\) and \(y\):
\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 = 7 \]
\[ (x - 1)^2 + (y - 1)^2 = 9 \]
Thus, the center of the first circle is \((1, 1)\) and the radius \(r_1 = 3\).

**For the second circle:**
Starting with the equation:
\[ 3(x^2 + y^2) - 8x + 29y = 0 \]
Dividing the entire equation by 3 gives:
\[ x^2 + y^2 - \frac{8}{3}x + \frac{29}{3}y = 0 \]
Rearranging it:
\[ x^2 + y^2 - \frac{8}{3}x + \frac{29}{3}y = 0 \]
Completing the square:
\[ \left(x - \frac{4}{3}\right)^2 - \frac{16}{9} + \left(y + \frac{29}{6}\right)^2 - \frac{841}{36} = 0 \]
Combining the constants:
\[ \left(x - \frac{4}{3}\right)^2 + \left(y + \frac{29}{6}\right)^2 = \frac{841}{36} + \frac{16}{9} \]
Finding a common denominator:
\[ \frac{841}{36} + \frac{64}{36} = \frac{905}{36} \]
Thus, the center of the second circle is \(\left(\frac{4}{3}, -\frac{29}{6}\right)\) and the radius \(r_2 = \sqrt{\frac{905}{36}} = \frac{\sqrt{905}}{6}\).

### Step 2: Calculate the distance between the centers
Let \(C_1 = (1, 1)\) and \(C_2 = \left(\frac{4}{3}, -\frac{29}{6}\right)\).

Using the distance formula:
\[ d = \sqrt{\left(\frac{4}{3} - 1\right)^2 + \left(-\frac{29}{6} - 1\right)^2} \]
Calculating:
\[ d = \sqrt{\left(\frac{1}{3}\right)^2 + \left(-\frac{35}{6}\right)^2} = \sqrt{\frac{1}{9} + \frac{1225}{36}} = \sqrt{\frac{4 + 1225}{36}} = \sqrt{\frac{1229}{36}} = \frac{\sqrt{1229}}{6} \]

### Step 3: Compare the distance with the sum of the radii
Now we compare the distance \(d\) with the sum of the radii \(r_1 + r_2\):
\[ r_1 + r_2 = 3 + \frac{\sqrt{905}}{6} \]
We need to check if \(d < r_1 + r_2\), \(d = r_1 + r_2\), or \(d > r_1 + r_2\).

### Step 4: Conclusion
If \(d < r_1 + r_2\), the circles intersect; if \(d = r_1 + r_2\), they touch externally; if \(d > r_1 + r_2\), they are separate.