Question

The number of common tangents, to the circles \( x^2 + y^2 - 18x - 15y + 131 = 0 \) and \( x^2 + y^2 - 6x - 6y - 7 = 0 \), is 

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is C

To determine the number of common tangents between the circles given by the equations:

• \(x^2 + y^2 - 18x - 15y + 131 = 0\) 
• \(x^2 + y^2 - 6x - 6y - 7 = 0\)

We first need to find the centers and radii of the circles.

Step 1: Rewrite the equations in standard form

The general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

For the first circle:

\(x^2 + y^2 - 18x - 15y + 131 = 0\)

Complete the square for \(x\) and \(y\):

• \(x^2 - 18x\): \((x - 9)^2 - 81\)
• \(y^2 - 15y\): \((y - 7.5)^2 - 56.25\)

Substitute back into the equation:

• \((x - 9)^2 - 81 + (y - 7.5)^2 - 56.25 + 131 = 0\)
• \((x - 9)^2 + (y - 7.5)^2 = 6.25\)

The center is \((9, 7.5)\) with a radius \(\sqrt{6.25} = 2.5\).

For the second circle:

\(x^2 + y^2 - 6x - 6y - 7 = 0\)

Complete the square for \(x\) and \(y\):

• \(x^2 - 6x\): \((x - 3)^2 - 9\)
• \(y^2 - 6y\): \((y - 3)^2 - 9\)

Substitute back into the equation:

• \((x - 3)^2 - 9 + (y - 3)^2 - 9 - 7 = 0\)
• \((x - 3)^2 + (y - 3)^2 = 25\)

The center is \((3, 3)\) with a radius \(\sqrt{25} = 5\).

Step 2: Determine the number of common tangents

The distance between the centers \((9, 7.5)\) and \((3, 3)\) is calculated as follows:

\(d = \sqrt{(9 - 3)^2 + (7.5 - 3)^2} = \sqrt{6^2 + 4.5^2} = \sqrt{36 + 20.25} = \sqrt{56.25} = 7.5\)

Since \(d = 7.5\) is greater than the sum of radii \((2.5 + 5 = 7.5)\) but not equal or overlapping, we check the conditions:

• If \(d > r_1 + r_2\), there are 4 common tangents.
• If \(d = r_1 + r_2\), there are 3 common tangents.

In this case, since \(d = r_1 + r_2\), there are 3 common tangents.

Conclusion

The number of common tangents to the given circles is 3.