Problem 12 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(x^{2} + y^{2} + 4x - 6y + 4 = 0\). We convert this into the standard form \((x-h)^2 + (y-k)^2 = r^2\) by completing the square.For \(x\), \((x^2 + 4x)\):- \(x^2 + 4x = (x+2)^2 - 4\).For \(y\), \((y^2 - 6y)\):- \(y^2 - 6y = (y-3)^2 - 9\).Substitute back: \((x + 2)^2 - 4 + (y - 3)^2 - 9 + 4 = 0\),\((x+2)^2 + (y-3)^2 = 9\).Thus, the center of the incircle is \((-2, 3)\) and the radius is 3.

Step 2: Relate the side length and radius of the incircle

For an equilateral triangle, the relation between the side length \(a\) and the inradius \(r\) is given by \(r = \frac{a\sqrt{3}}{6}\). Since we know \(r = 3\), we can find\[ a = \frac{6r}{\sqrt{3}} = 6 \times \frac{3}{\sqrt{3}} = 6\sqrt{3}. \]

Step 3: Calculate the circumradius of the equilateral triangle

For an equilateral triangle, the circumradius \(R\) is related to the side length \(a\) by the formula \( R = \frac{a}{\sqrt{3}} \). Substituting the side length \(a = 6\sqrt{3}\),\[ R = \frac{6\sqrt{3}}{\sqrt{3}} = 6. \]

Step 4: Find the center of the circumcircle

The center of the circumcircle of an equilateral triangle is the same as its centroid, which is the average of the vertices' coordinates. Since the triangle is symmetric and positioned around the incircle centered at \((-2, 3)\), the center of the circumcircle is the same, \((-2, 3)\).

Step 5: Write the equation of the circumcircle

The equation of the circumcircle, with center \((-2, 3)\) and radius 6, follows the standard form: \[(x+2)^2 + (y-3)^2 = 36. \]Expanding, we obtain \(x^2 + 4x + 4 + y^2 - 6y + 9 = 36\), hence \(x^2 + y^2 + 4x - 6y + 13 = 36\), which simplifies to\[ x^2 + y^2 + 4x - 6y - 23 = 0. \]

Step 6: Match with provided options

The equation \(x^2 + y^2 + 4x - 6y - 23 = 0\) matches option (B) as provided in the problem.

Key Concepts

Incircle of a Triangle

The incircle of a triangle is the largest circle that fits entirely within the triangle, touching all three sides at their respective tangents. This touching point is called the point of tangency.
The center of the incircle is known as the incenter. It is the point where the angle bisectors of the triangle intersect.

• The incenter is equidistant from all three sides of the triangle, making it the ideal center for the incircle.
• For any given triangle with a known inradius, the center coordinates and radius can be used to find the precise equation of the incircle in coordinate geometry.

In this specific exercise, the incircle's center is identified at (-2, 3), and its radius is found to be 3 based on the given incircle equation. The equation of an incircle in the standard form, \((x-h)^2 + (y-k)^2 = r^2\), accurately describes the circle within its triangle.

Equilateral Triangle properties

An equilateral triangle has some unique properties that differentiate it from other triangles. All sides of an equilateral triangle are equal, and each angle measures 60 degrees.

• The vertices of an equilateral triangle lie perfectly symmetrical around the center of its incircle and circumcircle.
• Formulas specific to equilateral triangles, such as those involving side lengths, inradius, and circumradius, can be particularly simple due to symmetry.

For instance, the side length \(a\) can determine the inradius \(r\) using \(r = \frac{a\sqrt{3}}{6}\), making everything in an equilateral triangle interdependent. Here, the side length derived from the inradius was used to compute the circumradius on the following basis: \( R = \frac{a}{\sqrt{3}} \).Equilateral triangles are often studied in exercises involving circles, as many geometric properties align and simplify calculations.

Coordinate Geometry

Coordinate geometry, or analytic geometry, involves using algebraic equations to represent geometric shapes in a coordinate plane. It allows for precise calculations and plotting of shapes such as circles and triangles.

• Equations of circles are derived and expanded in terms of \(x\) and \(y\), which represent any point on the circle.
• Using coordinates, properties of various geometric figures such as midpoints, centroids, inradii, and circumradii can be precisely calculated.

The exercise involves finding the equation of a circumcircle using coordinate geometry. The circumcircle has the same center as the incircle and a radius calculated through known properties of an equilateral triangle. Ultimately, these calculations allow one to derive the equation \(x^2 + y^2 + 4x - 6y - 23 = 0\), matching to one of the given options. This demonstrates how coordinate geometry seamlessly integrates geometric principles for problem-solving.