Problem 12 - A Complete Resource Book in Mathematics for JEE Main

Question

If the equation of the incircle of an equilateral triangle is \(x^{2}+y^{2}+4 x-6 y+4=0\), then the equation of the circumcircle of the triangle is (A) \(x^{2}+y^{2}+4 x+6 y-23=0\) (B) \(x^{2}+y^{2}+4 x-6 y-23=0\) (C) \(x^{2}+y^{2}-4 x-6 y-23=0\) (D) none of these

Step-by-Step Solution

Answer

The equation of the circumcircle is option (B): \(x^2 + y^2 + 4x - 6y - 23 = 0\).

Step 1: Identify the Center and Radius of the Incircle

The given equation of the incircle is \(x^{2} + y^{2} + 4x - 6y + 4 = 0\). To find the center and radius, we need to rewrite this equation in the standard form of a circle, \((x-h)^2 + (y-k)^2 = r^2\). Start by completing the square for the variables \(x\) and \(y\).

Step 2: Complete the Square for x-terms

The \(x\) terms in the equation are \(x^2 + 4x\). Complete the square: \(x^2 + 4x\) can be written as \((x+2)^2 - 4\).

Step 3: Complete the Square for y-terms

The \(y\) terms in the equation are \(y^2 - 6y\). Complete the square: \(y^2 - 6y\) can be written as \((y-3)^2 - 9\).

Step 4: Rewrite the Equation in Standard Form

Substitute the completed squares back into the equation:\( (x+2)^2 - 4 + (y-3)^2 - 9 + 4 = 0\).Simplify this to get \((x+2)^2 + (y-3)^2 = 9\).Thus, the center of the incircle is \((-2, 3)\) and the radius is \(3\).

Step 5: Find the Side of the Equilateral Triangle

For an equilateral triangle, the distance from the incenter \(r = \frac{a}{2 \sqrt{3}}\), where \(a\) is the side length. From the incircle, \(r = 3\), therefore:\[ 3 = \frac{a}{2 \sqrt{3}} \]Solving for \(a\) yields \(a = 6\sqrt{3}\).

Step 6: Calculate the Radius of the Circumcircle

The radius \(R\) of the circumcircle is related to the side length by \(R = \frac{a}{\sqrt{3}}\).Substitute \(a = 6\sqrt{3}\):\[ R = \frac{6\sqrt{3}}{\sqrt{3}} = 6 \]

Step 7: Determine the Equation of the Circumcircle

The circumcenter of an equilateral triangle is the same as the incenter, so it has the same center \((-2, 3)\). The radius of the circumcircle is \(6\).Thus, the equation of the circumcircle is:\((x+2)^2 + (y-3)^2 = 36\).

Step 8: Expand and Simplify the Circumcircle Equation

Expand: \[(x+2)^2 + (y-3)^2 = 36\]\[x^2 + 4x + 4 + y^2 - 6y + 9 = 36\]Combine and simplify:\[x^2 + y^2 + 4x - 6y - 23 = 0\]

Step 9: Compare with Given Options

The expanded and simplified equation of the circumcircle is \(x^2 + y^2 + 4x - 6y - 23 = 0\), which matches option (B).

Key Concepts

Incircle

An incircle of a triangle is a circle that fits snugly inside the triangle. It touches each side of the triangle exactly once. This means that the triangle's three sides are tangents to the incircle. The center of this circle is called the incenter, and it is the point where the bisectors of the triangle's angles meet.
The equation describing an incircle is typically presented in the form \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius of the circle. To find these, one usually completes the square for the given equation terms, as demonstrated in the solution above.

• In the given exercise, the incircle was identified with center \((-2, 3)\) and radius \(3\).
• The radius is sometimes linked to the side length of an equilateral triangle, expressed through the formula \(r = \frac{a}{2 \sqrt{3}}\).

This incenter also serves as the centroid when working with equilateral triangles, making these calculations especially crucial.

Circumcircle

The circumcircle of a triangle is the circle that passes through all three of the triangle's vertices. It encircles the entire triangle, hence the name circumcircle. The center of this circle is called the circumcenter.
For an equilateral triangle, the incenter and circumcenter both have identical positions, making calculations simpler. The radius of a circumcircle can be determined by the formula \(R = \frac{a}{\sqrt{3}}\), where \(a\) is the side length of the equilateral triangle. In the exercise, the circumcircle had a center at \((-2, 3)\) and a radius of 6.

• In our exercise, the circumcircle equation was simplified to \(x^2 + y^2 + 4x - 6y - 23 = 0\).
• Finding the right option required expanding and simplifying the equation until it matched one of the provided options.

The relationship between unit circle calculations and simpler algebraic representations is highlighted here, especially relevant in scenario-based problems like this.

Equilateral Triangle

An equilateral triangle is a special kind of triangle where all three sides are equal in length and all three angles measure 60 degrees. This symmetric structure leads to some unique properties involving circles.
For equilateral triangles, both the incircle and circumcircle share the same center point. This is because the geometry of equilateral triangles places all significant central points like the incenter, centroid, orthocenter, and circumcenter at the same spot.

• The side length \(a\) is crucial and affects both circle calculations: \(r = \frac{a}{2 \sqrt{3}}\) for the incircle, and \(R = \frac{a}{\sqrt{3}}\) for the circumcircle.
• In our exercise, the side length was found to be \(6\sqrt{3}\), derived from the relation involving the incircle's radius.

Understanding such core properties of equilateral triangles simplifies calculations and offers deeper insights into equations involving geometrical locations and dimensions.