If the area of an equilateral triangle inscribed in the circle, x² + y² + 10x + 12y + c = 0 is 27√3

Q: Wait, why is the radius formula $R = \sqrt{g^2 + f^2 - c}$? Shouldn't it be plus $c$?

That is a classic trap! Remember the standard form of a circle is $(x-h)^2 + (y-k)^2 = R^2$. When you expand $(x+g)^2 + (y+f)^2 = R^2$, you get $x^2 + 2gx + g^2 + y^2 + 2fy + f^2 = R^2$, which rearranges to $x^2 + y^2 + 2gx + 2fy + (g^2 + f^2 - R^2) = 0$. Comparing this to our given equation, $c = g^2 + f^2 - R^2$. Solving for $R^2$ gives $R^2 = g^2 + f^2 - c$, so $R = \sqrt{g^2 + f^2 - c}$. Always keep an eye on that constant term!

Q: How do we know for sure that $a = R\sqrt{3}$ for an equilateral triangle?

Great question! In an equilateral triangle, the circumcenter is also the centroid. The distance from the centroid to a vertex is the circumradius $R$. If you draw a perpendicular from the center to a side, it bisects the side (length $a/2$) and forms a $30-60-90$ triangle with the radius $R$ and the segment $a/2$. Using trigonometry, $\cos(30^\circ) = \frac{a/2}{R}$, so $\frac{\sqrt{3}}{2} = \frac{a}{2R}$. Simplifying this gives $a = R\sqrt{3}$. It is a beautiful geometric result that saves you a lot of time in exams!

Q: Can I solve this using the distance from the center to the side of the triangle?

Absolutely! That is a brilliant alternative approach. The distance from the center to any side of an equilateral triangle is the apothem, $r = R \sin(30^\circ) = R/2$. You could use the area formula $A = \frac{1}{2} \times \text{perimeter} \times \text{apothem}$ or simply use the geometry of the triangle to relate the height to the radius. However, the $a = R\sqrt{3}$ relation is usually the fastest path. Keep exploring these different methods; it strengthens your intuition!

Question

If the area of an equilateral triangle inscribed in the circle, $x^2 + y^2 + 10x + 12y + c = 0$ is $27\sqrt{3}$ sq. units then c is equal to :

Options:

(A) 20

(B) 25

(C) 13

(D) -25

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: Unlocking the Circle

Imagine you are standing on a vast, empty plane. Before you, a perfect circle is etched into the ground, defined by the equation $x^2 + y^2 + 10x + 12y + c = 0$.

Inside this circle, a perfectly symmetric equilateral triangle is inscribed, its vertices kissing the boundary of the circle. This is not just a problem of algebra; it is a dance of geometry.

Our goal is to find the hidden constant $c$, the value that dictates the very size of this circle. Let us peel back the layers of this problem together.

### Phase 1: Decoding the Circle

Every circle has a soul, defined by its center and its radius. The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.

By comparing this to our given equation, $x^2 + y^2 + 10x + 12y + c = 0$, we can immediately extract the parameters. We see that $2g = 10$, which implies $g = 5$, and $2f = 12$, which implies $f = 6$.

These values, $g$ and $f$, are the keys to the kingdom. They allow us to express the radius $R$ in terms of our unknown $c$ using the standard formula:

$$R = \sqrt{g^2 + f^2 - c}$$

Substituting our values, we obtain:

$$R = \sqrt{5^2 + 6^2 - c} = \sqrt{25 + 36 - c} = \sqrt{61 - c}$$

Squaring both sides, we find our anchor equation:

$$R^2 = 61 - c$$

### Phase 2: The Triangle's Secret

The problem gifts us with the area of the inscribed equilateral triangle: $27\sqrt{3}$ square units. We know the area $A$ of an equilateral triangle with side length $a$ is given by the formula:

$$A = \frac{\sqrt{3}}{4}a^2$$

By setting this equal to the given area, we have:

$$\frac{\sqrt{3}}{4}a^2 = 27\sqrt{3}$$

The $\sqrt{3}$ terms on both sides cancel out, leaving us with:

$$\frac{a^2}{4} = 27 \Rightarrow a^2 = 108$$

Taking the square root, we find $a = \sqrt{108} = 6\sqrt{3}$. We have successfully bridged the gap between the area and the side length.

### Phase 3: The Geometric Bridge

Now, we must connect the triangle to the circle. For an equilateral triangle inscribed in a circle, there is a fundamental relationship between the side length $a$ and the circumradius $R$:

$$a = R\sqrt{3}$$

Substituting our value for $a$, we get:

$$6\sqrt{3} = R\sqrt{3}$$

The $\sqrt{3}$ terms cancel out, revealing that the radius $R$ is exactly $6$ units.

### Phase 4: The Final Revelation

We have arrived at the final step. We know $R = 6$, which means $R^2 = 36$.

We also know from our earlier work that $R^2 = 61 - c$. Substituting $36$ for $R^2$, we get:

$$36 = 61 - c$$

Rearranging this linear equation, we find:

$$c = 61 - 36$$

The mystery is solved. The value of the constant is **$c = 25$**.

Animated Solution for Mathematics - Circles: If the area of an equilateral triangle inscribed in the circle, $x^{2} + y^{2} + 10 x + 12 y + c = 0$ is $273$ sq. units then c is equal to :

Select Answer:

Visualized Solution

Visualizing the Problem

Extracting Circle Parameters

Formulating the Radius

Area of the Equilateral Triangle

Setting up the Equation

Calculating the Side Length

Relating Side $a$ and Radius $R$

Calculating the Circumradius

Finding the Unknown $c$

The Final Value

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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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; } If the area of an equilateral triangle inscribed in the circle, x2+y2+10x+12y+c=0 is 273​ sq. units then c is equal to :