Problem 63 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If two vertices of an equilateral triangle are \(\mathrm{A}(-a, 0)\) and \(B(a, 0), a>0\), and the third vertex \(\mathrm{C}\) lies above \(x\)-axis then the equation of the circumcircle of \(\triangle \mathrm{ABC}\) is: (a) \(3 x^{2}+3 y^{2}-2 \sqrt{3} a y=3 a^{2}\) (b) \(3 x^{2}+3 y^{2}-2 a y=3 a^{2}\) (c) \(x^{2}+y^{2}-2 a y=a^{2}\) (d) \(x^{2}+y^{2}-\sqrt{3} a y=a^{2}\)

Step-by-Step Solution

Answer

The equation of the circumcircle is (a) 3x^2 + 3y^2 - 2√3 ay = 3a^2.

Step 1: Find Coordinates of Vertex C

Since the triangle is equilateral and points A and B are (-a, 0) and (a, 0) respectively, the third point C will lie on a perpendicular from the midpoint of AB with a distance a from this midpoint. First, find the midpoint M of AB, which is (0,0). Then, use the fact that ABC is equilateral to determine the height from midpoint M to C:The height h of an equilateral triangle is \( h = \sqrt{3}/2 \times \text{side length} \).The side length here is 2a, so the height h = \( \sqrt{3}a \). Thus, C = \((0, \sqrt{3}a)\).

Step 2: Determine Circumcenter and Radius

The circumcenter of any triangle is the center of the circle that passes through all three vertices. For an equilateral triangle, the circumcenter coincides with the centroid because all medians and altitudes are equal. The centroid G of \(\triangle ABC\) can be computed as the average of the vertex coordinates: \( G = \left(\frac{-a+a+0}{3}, \frac{0+0+\sqrt{3}a}{3}\right) = \left(0, \frac{\sqrt{3}a}{3}\right) \). The circumradius R for an equilateral triangle can be found using the formula \( R = \frac{\text{side length}}{\sqrt{3}} = \frac{2a}{\sqrt{3}} = \frac{2a}{\sqrt{3}} \).

Step 3: Formulate Equation of Circumcircle

The equation of a circle with center at \((h,k)\) and radius R is \((x-h)^2 + (y-k)^2 = R^2 \). Substitute \(h=0\), \(k= \frac{\sqrt{3}a}{3}\), and \(R = \frac{2a}{\sqrt{3}}\) into the equation:\[(x-0)^2 + \left(y - \frac{\sqrt{3}a}{3}\right)^2 = \left(\frac{2a}{\sqrt{3}}\right)^2 \]which simplifies to:\[x^2 + y^2 - 2\cdot\frac{\sqrt{3}a}{3}\cdot y + \left(\frac{\sqrt{3}a}{3}\right)^2 = \frac{4a^2}{3}\]Simplify further to obtain the equation.Combine all terms:\[x^2 + y^2 - \frac{2\sqrt{3}ay}{3} + \frac{a^2}{3} = \frac{4a^2}{3} \]Finally, simplify to match one of the provided choices.

Step 4: Simplify and Compare with Options

Simplify the expression obtained in the previous step.Multiply the whole equation by 3 to remove fractions:\[3(x^2 + y^2 - \frac{2\sqrt{3}ay}{3} + \frac{a^2}{3}) = 3\cdot\frac{4a^2}{3}\]which simplifies to:\[3x^2 + 3y^2 - 2\sqrt{3}ay + a^2 = 4a^2\].Rearranging gives:\[3x^2 + 3y^2 - 2\sqrt{3}ay = 3a^2\]This matches option (a), which is the correct equation for the circumcircle.

Key Concepts

Circumcircle

In geometry, the circumcircle is the circle that passes through all three vertices of a triangle. It's a key concept, as it provides a simple way to understand the triangle's properties.
The center of the circumcircle is the circumcenter. For an equilateral triangle, like \( \triangle ABC \), the circumcenter is special. It coincides with the centroid, meaning the point where all medians intersect is also the circumcenter.

• The circumradius is the radius of the circumcircle. In equilateral triangles, it's related directly to the side length. The formula is \( R = \frac{\text{side length}}{\sqrt{3}} \).
• Having a circumradius helps us write the equation of the circumcircle. Given the circumcenter and circumradius, the circle's equation is \( (x-h)^2 + (y-k)^2 = R^2 \), where \( (h, k) \) are the center's coordinates.

These relationships make working with equilateral triangles and circumcircles an elegant exercise.

Coordinate Geometry

Coordinate geometry, or analytic geometry, combines algebra and geometry to solve problems involving shapes and their properties.
It uses a coordinate system, like the Cartesian plane, to plot points and visualize figures such as triangles.

• With two given vertices of an equilateral triangle located at \( A(-a, 0) \) and \( B(a, 0) \), coordinate geometry helps locate the third vertex \( C \).
• By creating a perpendicular from the midpoint of \( AB \) and using the known side length, we can determine \( C's \) coordinates. For equilateral triangles, this involves calculating the height, which is \( \frac{\sqrt{3}}{2} \) times the side length.
• This approach simplifies finding essential points like the centroid or circumcenter.

Coordinate geometry not only aids in finding points but also helps formulate various equations, such as that of a circle.

Centroid

The centroid of a triangle is the point where all its medians intersect. It's a point of balance in a triangle, often described as the triangle's "center of mass."
In \( \triangle ABC \), the centroid can be easily calculated using the average of its vertices' coordinates.

• For example, the centroid \( G \) of the triangle with vertices \( (-a,0) \), \( (a,0) \), and \( (0, \sqrt{3}a) \) is found with the formula: \( G = \left(\frac{-a+a+0}{3}, \frac{0+0+\sqrt{3}a}{3}\right) = \left(0, \frac{\sqrt{3}a}{3}\right) \).
• In equilateral triangles, the centroid, circumcenter, and orthocenter all coincide, simplifying many geometric operations.

Understanding the centroid offers insight into the triangle's symmetry and balance. It's not just a midpoint, but a pivotal point in triangle geometry.