The common tangent to the circles x² + y² = 4 and x² + y² + 6x + 8y - 24 = 0 also passes through the

Q: Wait, why do we subtract the equations to get the tangent? Can't we just use the slope-point form?

You absolutely could use the slope-point form, but that would require you to solve for the slope $m$ using the condition of tangency, which often leads to a messy quadratic equation. The radical axis method is a powerful geometric shortcut specifically for touching circles. It is elegant, efficient, and saves you from unnecessary algebraic heavy lifting.

Q: How do I know if the circles touch internally or externally?

It all comes down to the distance between the centers $d$. If $d = r_1 + r_2$, the circles touch externally. If $d = |r_1 - r_2|$, they touch internally. In this problem, $d = 5$ and $|7 - 2| = 5$. Since they are equal, we know for certain they touch internally.

Q: What if the circles didn't touch? Would the radical axis still be the common tangent?

No, it wouldn't! The radical axis is the locus of points with equal power with respect to both circles. It is always a straight line, but it only acts as the common tangent when the circles are tangent to each other. If the circles intersect at two distinct points, the radical axis is simply the common chord.

Question

The common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 6x + 8y - 24 = 0$ also passes through the point :-

Options:

(A) $(-4, 6)$

(B) $(6, -2)$

(C) $(-6, 4)$

(D) $(4, -2)$

Sigma Percentile · Accepted Answer

### Analyzing the Circles

First, let's define our players. The first circle, $S_1$, is defined by $x^2 + y^2 = 4$. This is a classic, centered at the origin $(0, 0)$ with a radius $r_1 = 2$.

Now, consider the second circle, $S_2$, given by $x^2 + y^2 + 6x + 8y - 24 = 0$. Using the general form of a circle, we identify its center $C_2$ as $(-3, -4)$.

To find its radius $r_2$, we use the formula $\sqrt{g^2 + f^2 - c}$:

$$r_2 = \sqrt{(-3)^2 + (-4)^2 - (-24)} = \sqrt{9 + 16 + 24} = \sqrt{49} = 7$$

So, we have a small circle of radius $2$ and a larger one of radius $7$.

### The Distance Between Centers

To understand how these circles interact, we calculate the distance $d$ between their centers $C_1(0, 0)$ and $C_2(-3, -4)$. Using the distance formula:

$$d = \sqrt{(-3 - 0)^2 + (-4 - 0)^2} = \sqrt{9 + 16} = 5$$

Now, look at the relationship between this distance and the radii. The difference between the radii is $|r_2 - r_1| = |7 - 2| = 5$.

The fact that $d = |r_2 - r_1|$ is a massive geometric signal. It tells us that the circles are not just near each other; they are **touching internally**. One circle is nestled perfectly inside the other, sharing a single point of contact.

### The Radical Axis Shortcut

Because the circles touch internally, they share exactly one common tangent at that point of contact. In the world of JEE, we love shortcuts, and this is a big one: for touching circles, the common tangent is the **radical axis**.

We find this by subtracting the equations of the two circles: $S_1 - S_2 = 0$. Let's set it up:

$$(x^2 + y^2 - 4) - (x^2 + y^2 + 6x + 8y - 24) = 0$$

Watch as the quadratic terms $x^2$ and $y^2$ vanish, leaving us with a beautiful linear equation:

$$-6x - 8y + 20 = 0$$

Dividing by $-2$, we get the equation of the tangent:

$$3x + 4y - 10 = 0 \quad 	ext{or} \quad 3x + 4y = 10$$

### The Final Verification

We have our line. Now, we just need to see which of the given points lies on it. Testing the point $(6, -2)$, we substitute $x = 6$ and $y = -2$ into $3x + 4y = 10$:

$$3(6) + 4(-2) = 18 - 8 = 10$$

It works perfectly! The left-hand side equals the right-hand side.

We have successfully navigated the geometry, identified the internal tangency, utilized the radical axis, and verified our result. The final equation of the common tangent is **$3x + 4y = 10$**.

Animated Solution for Mathematics - Circles: The common tangent to the circles $x^{2} + y^{2} = 4$ and $x^{2} + y^{2} + 6 x + 8 y - 24 = 0$ also passes through the point :-

Select Answer:

Visualized Solution

Analyze Circle $S_{1}$

Analyze Circle $S_{2}$

Radius of Circle $S_{2}$

Distance Between Centers

Determine Relative Position

Common Tangent Concept

Setup Tangent Equation

Simplify Tangent Equation

Final Tangent Equation

Verify the Options

Conclusion

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Circles

The Distance Between Centers

The Radical Axis Shortcut

The Final Verification

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The common tangent to the circles x2+y2=4 and x2+y2+6x+8y−24=0 also passes through the point :-