Master applications of circles in CBSE Class 11 Applied Mathematics. Learn to find equations, radii, and centers with solved examples and practice sets.
A circle is the locus of a point which moves in a plane such that its distance from a fixed point (center) is always constant (radius). The standard form of a circle with center (h, k) and radius r is derived from the distance formula. This equation is fundamental for solving problems involving circular paths or boundaries.
Step 1: Identify center (h, k) = (2, -3) and radius r = 5.
Step 2: Substitute into (x - h)² + (y - k)² = r².
Step 3: (x - 2)² + (y + 3)² = 25.
Answer: x² + y² - 4x + 6y - 12 = 0.
The general form represents a circle when the coefficients of x² and y² are equal and there is no xy term. By completing the square, we can transform the general equation into the standard form to identify the center and radius. This is essential for analyzing circular motion or geometric constraints.
Step 1: Given x² + y² + 4x - 6y - 3 = 0.
Step 2: Compare with 2g = 4 and 2f = -6, so g = 2, f = -3.
Step 3: Center is (-g, -f) = (-2, 3). Radius is √(g² + f² - c) = √(4 + 9 + 3) = 4.
Answer: Center (-2, 3), Radius 4.
To find the equation of a circle passing through three non-collinear points, we substitute each point (x, y) into the general equation x² + y² + 2gx + 2fy + c = 0. This results in a system of three linear equations in variables g, f, and c. Solving this system allows us to define the unique circle passing through these points.
Step 1: (0,0) gives c = 0.
Step 2: (2,0) gives 4 + 4g = 0 → g = -1.
Step 3: (0,2) gives 4 + 4f = 0 → f = -1.
Answer: x² + y² - 2x - 2y = 0.