Question:
The equation of the common tangent to the circles $x^2 + y^2 - 4x + 10y + 20 = 0$ and $x^2 + y^2 + 8x - 6y - 24 = 0$ is
The equation of the common tangent to the circles $x^2 + y^2 - 4x + 10y + 20 = 0$ and $x^2 + y^2 + 8x - 6y - 24 = 0$ is
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Whenever a question asks for a common tangent of two circles, start by checking $S_1 - S_2 = 0$. If that linear combination matches one of the options perfectly, it is almost certainly the correct tangent line at the contact interface!
Updated On: Jun 12, 2026
- $3x - 4y + 11 = 0$
- $3x - 4y - 11 = 0$
- $3x + 4y + 11 = 0$
- $3x + 4y - 11 = 0$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We are given the equations of two circles and need to find the equation of their common tangent line.
Step 2: Key Formula or Approach:
First, find the centers ($C_1, C_2$) and radii ($r_1, r_2$) of both circles. Compute the distance between their centers ($C_1C_2$).
If $C_1C_2 = r_1 + r_2$, the two circles touch each other externally.
When two circles touch externally, the common tangent at their point of contact is the same as their radical axis, which is found by subtracting the two circle equations: $S_1 - S_2 = 0$.
Step 3: Detailed Explanation:
Let's find the centers and radii of each circle:
1. For Circle $S_1: x^2 + y^2 - 4x + 10y + 20 = 0$:
Center $C_1 = \left(-\frac{g}{2}, -\frac{f}{2}\right) = (2, -5)$
Radius $r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (5)^2 - 20} = \sqrt{4 + 25 - 20} = \sqrt{9} = 3$
2. For Circle $S_2: x^2 + y^2 + 8x - 6y - 24 = 0$:
Center $C_2 = (-4, 3)$
Radius $r_2 = \sqrt{(4)^2 + (-3)^2 - (-24)} = \sqrt{16 + 9 + 24} = \sqrt{49} = 7$
Now calculate the distance between the two center points $C_1(2, -5)$ and $C_2(-4, 3)$:
$$C_1C_2 = \sqrt{(-4 - 2)^2 + (3 - (-5))^2} = \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$ Notice that:
$$r_1 + r_2 = 3 + 7 = 10$$ Since $C_1C_2 = r_1 + r_2$, the circles touch each other externally. Thus, the common tangent equation is given directly by $S_1 - S_2 = 0$:
$$(x^2 + y^2 - 4x + 10y + 20) - (x^2 + y^2 + 8x - 6y - 24) = 0$$ $$-4x - 8x + 10y - (-6y) + 20 - (-24) = 0$$ $$-12x + 16y + 44 = 0$$ Divide the entire equation by $-4$ to simplify:
$$3x - 4y - 11 = 0$$ This matches option (B).
Step 4: Final Answer:
The equation of the common tangent is $3x - 4y - 11 = 0$, which corresponds to option (B).
We are given the equations of two circles and need to find the equation of their common tangent line.
Step 2: Key Formula or Approach:
First, find the centers ($C_1, C_2$) and radii ($r_1, r_2$) of both circles. Compute the distance between their centers ($C_1C_2$).
If $C_1C_2 = r_1 + r_2$, the two circles touch each other externally.
When two circles touch externally, the common tangent at their point of contact is the same as their radical axis, which is found by subtracting the two circle equations: $S_1 - S_2 = 0$.
Step 3: Detailed Explanation:
Let's find the centers and radii of each circle:
1. For Circle $S_1: x^2 + y^2 - 4x + 10y + 20 = 0$:
Center $C_1 = \left(-\frac{g}{2}, -\frac{f}{2}\right) = (2, -5)$
Radius $r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{(-2)^2 + (5)^2 - 20} = \sqrt{4 + 25 - 20} = \sqrt{9} = 3$
2. For Circle $S_2: x^2 + y^2 + 8x - 6y - 24 = 0$:
Center $C_2 = (-4, 3)$
Radius $r_2 = \sqrt{(4)^2 + (-3)^2 - (-24)} = \sqrt{16 + 9 + 24} = \sqrt{49} = 7$
Now calculate the distance between the two center points $C_1(2, -5)$ and $C_2(-4, 3)$:
$$C_1C_2 = \sqrt{(-4 - 2)^2 + (3 - (-5))^2} = \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$ Notice that:
$$r_1 + r_2 = 3 + 7 = 10$$ Since $C_1C_2 = r_1 + r_2$, the circles touch each other externally. Thus, the common tangent equation is given directly by $S_1 - S_2 = 0$:
$$(x^2 + y^2 - 4x + 10y + 20) - (x^2 + y^2 + 8x - 6y - 24) = 0$$ $$-4x - 8x + 10y - (-6y) + 20 - (-24) = 0$$ $$-12x + 16y + 44 = 0$$ Divide the entire equation by $-4$ to simplify:
$$3x - 4y - 11 = 0$$ This matches option (B).
Step 4: Final Answer:
The equation of the common tangent is $3x - 4y - 11 = 0$, which corresponds to option (B).
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