Question

The number of common tangents to the circles $x^2 + y^2 - 6x - 14y + 48 = 0$ and $x^2 + y^2 = 6x$ is :

• A. 4
• B. 1
• C. 2
• D. 0

Hint:

Remember the common tangent rules based on distance $d$:
- $d > r_1+r_2 \implies 4$ common tangents.
- $d = r_1+r_2 \implies 3$ common tangents (touch externally).
- $|r_1-r_2| < d < r_1+r_2 \implies 2$ common tangents (intersect).
- $d = |r_1-r_2| \implies 1$ common tangent (touch internally).
- $d < |r_1-r_2| \implies 0$ common tangents.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The number of common tangents between two circles is determined by comparing the distance $d$ between their centers with the sum and difference of their radii.

Step 2: Detailed Explanation:
Let us find the centers and radii of both circles:
1. For Circle 1 ($C_1$):
\[ x^2 + y^2 - 6x - 14y + 48 = 0 \]
Center, $O_1 = (3, 7)$
Radius, $r_1 = \sqrt{3^2 + 7^2 - 48} = \sqrt{9 + 49 - 48} = \sqrt{10} \approx 3.16$
2. For Circle 2 ($C_2$):
\[ x^2 + y^2 - 6x = 0 \]
Center, $O_2 = (3, 0)$
Radius, $r_2 = \sqrt{3^2 + 0^2 - 0} = \sqrt{9} = 3$
Calculate the distance $d$ between the centers $O_1$ and $O_2$:
\[ d = \sqrt{(3-3)^2 + (7-0)^2} = \sqrt{0 + 49} = 7 \]
Now, calculate the sum of their radii:
\[ r_1 + r_2 = \sqrt{10} + 3 \approx 3.16 + 3 = 6.16 \]
Let us compare the distance $d$ with the sum of the radii:
\[ d = 7 > r_1 + r_2 \approx 6.16 \]
Since the distance between their centers is strictly greater than the sum of their radii, the two circles lie completely outside each other.
Two non-intersecting, separate circles have exactly 4 common tangents (2 direct common tangents and 2 transverse common tangents).
Therefore, the number of common tangents is 4.

Step 3: Final Answer
The correct option is (A).