RD Sharma Solutions for Class 11 Maths Chapter 24 - Free PDF Download
FAQs on RD Sharma Class 11 Maths Solutions Chapter 24 - The Circle
1. What key topics are covered in Vedantu’s RD Sharma Solutions for Class 11 Maths Chapter 24, The Circle?
The RD Sharma Solutions for Class 11 Maths Chapter 24 provide comprehensive, step-by-step methods for all topics aligned with the CBSE syllabus. Key areas covered include:
- Finding the standard and general equations of a circle.
- Determining the centre and radius from a given equation.
- Solving for the equation of a circle under various conditions (e.g., passing through three points, given endpoints of a diameter).
- Understanding the parametric form of a circle.
- Analysing the position of a point or a line with respect to a circle.
- Calculating the equation and length of a tangent and normal to a circle.
2. What is the standard method shown in the solutions for finding the equation of a circle given its centre and radius?
The solutions demonstrate a clear, step-by-step method. First, identify the coordinates of the centre, let's say (h, k), and the length of the radius, r. Then, apply the standard form of the circle's equation, which is (x - h)² + (y - k)² = r². The final step involves substituting the values of h, k, and r into this formula and simplifying the expression to get the required equation.
3. How do the RD Sharma solutions explain finding the centre and radius from the general equation of a circle?
The solutions provide a systematic approach. The given equation is first compared to the general form x² + y² + 2gx + 2fy + c = 0. By comparing coefficients, you can find the values of g, f, and c. The centre of the circle is then determined using the formula (-g, -f), and the radius is calculated using the formula r = √(g² + f² - c). This method is shown with multiple examples to ensure clarity.
4. Why is it crucial to check the condition g² + f² - c > 0 when finding the radius from the general equation of a circle?
This condition is a critical checkpoint that the solutions emphasise. The expression g² + f² - c represents the square of the radius. For a real circle to exist, its radius must be a positive real number.
- If g² + f² - c > 0, the equation represents a real circle.
- If g² + f² - c = 0, the radius is zero, and the equation represents a point circle.
- If g² + f² - c < 0, the radius would be an imaginary number, meaning no real circle can be drawn. The solutions highlight this to prevent incorrect interpretations of the geometric figure.
5. How do the solutions approach problems where the equation of a circle passing through three non-collinear points is required?
The solutions break this complex problem down into manageable steps. The standard approach involves assuming the general equation of the circle (x² + y² + 2gx + 2fy + c = 0). Since all three given points lie on the circle, their coordinates must satisfy the equation. By substituting the (x, y) coordinates of each of the three points, a system of three linear equations in g, f, and c is formed. The solutions then demonstrate how to solve this system simultaneously to find the unique values for g, f, and c, which are then substituted back into the general equation.
6. What is a common mistake students make when determining the position of a point relative to a circle, and how do the solutions address it?
A common mistake is simply comparing distances without a systematic formula. The RD Sharma solutions teach the correct method using the expression S₁ = x₁² + y₁² + 2gx₁ + 2fy₁ + c, where (x₁, y₁) is the point. The solutions clarify that:
- If S₁ > 0, the point lies outside the circle.
- If S₁ = 0, the point lies on the circle.
- If S₁ < 0, the point lies inside the circle. This algebraic method is faster and less prone to error than calculating the distance from the centre and comparing it with the radius, a pitfall the solutions help students avoid.
7. How do the RD Sharma solutions differentiate the method for finding a tangent at a point on the circle versus from an external point?
The solutions clearly distinguish between these two scenarios. For a tangent at a given point (x₁, y₁) on the circle, a direct substitution method is used. The equation is derived by replacing x² with xx₁, y² with yy₁, 2x with (x+x₁), and 2y with (y+y₁). However, for finding tangents from an external point, the method involves assuming the tangent's equation as y = mx + c, using the condition of tangency (the perpendicular distance from the centre to this line equals the radius), and solving for the slope 'm'. This typically results in a quadratic equation, yielding two distinct tangents.
