Question:medium

The number of common tangents that can be drawn to the circles $x^2 + y^2 - 6x = 0$ and $x^2 + y^2 + 6x + 2y + 1 = 0$ is ________

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Intersecting circles → 3 common tangents.
Updated On: May 14, 2026
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
To ascertain the number of common tangents to a pair of circles, we must determine their geometrical relative position. This is achieved by calculating the distance $d$ between their centers and comparing it against the sum ($r_1 + r_2$) and the absolute difference ($|r_1 - r_2|$) of their radii. Step 2: Key Formula or Approach:
1. Center and radius of general circle $x^2 + y^2 + 2gx + 2fy + c = 0$ are $(-g, -f)$ and $\sqrt{g^2 + f^2 - c}$. 2. If $d>r_1 + r_2$, circles are strictly outside (disjoint) $\rightarrow 4$ common tangents. 3. If $d = r_1 + r_2$, circles touch externally $\rightarrow 3$ common tangents. 4. If $|r_1 - r_2|<d<r_1 + r_2$, circles intersect $\rightarrow 2$ common tangents. 5. If $d = |r_1 - r_2|$, circles touch internally $\rightarrow 1$ common tangent. 6. If $d<|r_1 - r_2|$, one circle is inside the other $\rightarrow 0$ common tangents. Step 3: Detailed Explanation:
Let's analyze the first circle, $C_1$: Equation: $x^2 + y^2 - 6x = 0$ Comparing with standard form, $2g = -6 \implies g = -3$; $2f = 0 \implies f = 0$; $c = 0$. Center $O_1 = (-g, -f) = (3, 0)$. Radius $r_1 = \sqrt{(-3)^2 + 0^2 - 0} = \sqrt{9} = 3$. Let's analyze the second circle, $C_2$: Equation: $x^2 + y^2 + 6x + 2y + 1 = 0$ Comparing with standard form, $2g = 6 \implies g = 3$; $2f = 2 \implies f = 1$; $c = 1$. Center $O_2 = (-g, -f) = (-3, -1)$. Radius $r_2 = \sqrt{3^2 + 1^2 - 1} = \sqrt{9} = 3$. Now compute the distance $d$ between centers $O_1(3, 0)$ and $O_2(-3, -1)$: \[ d = \sqrt{(-3 - 3)^2 + (-1 - 0)^2} \] \[ d = \sqrt{(-6)^2 + (-1)^2} \] \[ d = \sqrt{36 + 1} = \sqrt{37} \] Next, calculate the sum of their radii: \[ r_1 + r_2 = 3 + 3 = 6 \] Now compare the distance $d$ with the sum of radii $r_1 + r_2$: We know $6 = \sqrt{36}$. Since $\sqrt{37}>\sqrt{36}$, it means $d>r_1 + r_2$. This condition signifies that the two circles are completely disjoint; they lie strictly outside each other without touching or intersecting. In this configuration, exactly 4 common tangents can be drawn (two direct common tangents and two transverse common tangents). Step 4: Final Answer:
The number of common tangents is 4.
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