What is the equation of the circle which passes through the points (3, -2) and (-2, 0) and having its centre on the line 2x – y – 3 = 0?
x 2+ y 2+ 3x + 12y + 2 = 0
Finding the Equation of a Circle
The problem asks for the equation of a circle that passes through two specific points, (3, -2) and (-2, 0), and has its center located on the line 2x – y – 3 = 0. To solve this, we can use the general equation of a circle and the given conditions to find the unknown coefficients.
General Equation of a Circle
The general equation of a circle is given by: \(x^2 + y^2 + 2gx + 2fy + c = 0\) In this equation, the coordinates of the center of the circle are (-g, -f), and the radius squared is \(r^2 = g^2 + f^2 - c\).
Using the Given Conditions
We are given two points that lie on the circle and a line that the center lies on. Let's use this information:
Condition 1: Center lies on the line 2x – y – 3 = 0
The center of the circle is (-g, -f). Since this point lies on the line 2x – y – 3 = 0, substituting the center coordinates into the line equation gives:
\(2(-g) - (-f) - 3 = 0\) \(-2g + f - 3 = 0\) \(f = 2g + 3\) (Equation A)This equation relates the coefficients 'f' and 'g'.
Condition 2: Circle passes through point (3, -2)
Since the point (3, -2) lies on the circle, it must satisfy the general equation of the circle. Substituting x=3 and y=-2:
\((3)^2 + (-2)^2 + 2g(3) + 2f(-2) + c = 0\) \(9 + 4 + 6g - 4f + c = 0\) \(13 + 6g - 4f + c = 0\) (Equation B)Condition 3: Circle passes through point (-2, 0)
Since the point (-2, 0) lies on the circle, it must satisfy the general equation of the circle. Substituting x=-2 and y=0:
\((-2)^2 + (0)^2 + 2g(-2) + 2f(0) + c = 0\) \(4 + 0 - 4g + 0 + c = 0\) \(4 - 4g + c = 0\) (Equation C)Solving the System of Equations
We now have a system of three linear equations with three unknowns (g, f, c):
- \(f = 2g + 3\)
- \(13 + 6g - 4f + c = 0\)
- \(4 - 4g + c = 0\)
Let's substitute Equation A into Equation B:
\(13 + 6g - 4(2g + 3) + c = 0\) \(13 + 6g - 8g - 12 + c = 0\) \(1 - 2g + c = 0\) (Equation D)Now we have a simpler system with two equations and two unknowns (g, c) from Equation C and Equation D:
- \(1 - 2g + c = 0\)
- \(4 - 4g + c = 0\)
Subtract Equation D from Equation C:
\((4 - 4g + c) - (1 - 2g + c) = 0\) \(4 - 4g + c - 1 + 2g - c = 0\) \(3 - 2g = 0\) \(2g = 3\) \(g = \frac{3}{2}\)Now substitute the value of g back into Equation D to find c:
\(1 - 2(\frac{3}{2}) + c = 0\) \(1 - 3 + c = 0\) \(-2 + c = 0\) \(c = 2\)Finally, substitute the value of g back into Equation A to find f:
\(f = 2g + 3\) \(f = 2(\frac{3}{2}) + 3\) \(f = 3 + 3\) \(f = 6\)The Equation of the Circle
We have found the values of the coefficients: \(g = \frac{3}{2}\), \(f = 6\), and \(c = 2\). Substitute these values back into the general equation of the circle \(x^2 + y^2 + 2gx + 2fy + c = 0\):
\(x^2 + y^2 + 2(\frac{3}{2})x + 2(6)y + 2 = 0\) \(x^2 + y^2 + 3x + 12y + 2 = 0\)This is the equation of the circle that satisfies all the given conditions.
Summary of Steps
Here is a brief overview of the process:
- Start with the general equation of a circle.
- Use the condition about the center lying on the line to get a relationship between 'g' and 'f'.
- Substitute the coordinates of the two given points into the general equation to get two more equations involving 'g', 'f', and 'c'.
- Solve the system of three linear equations to find the values of 'g', 'f', and 'c'.
- Substitute these values back into the general equation to get the final equation of the circle.
| Coefficient | Value |
|---|---|
| g | \(\frac{3}{2}\) |
| f | 6 |
| c | 2 |
Revision Table: Circle Equation Concepts
| Concept | Description | Formula/Representation |
|---|---|---|
| General Equation of Circle | Standard form used for calculations involving points and center. | \(x^2 + y^2 + 2gx + 2fy + c = 0\) |
| Center of Circle (from General Eq) | Coordinates of the circle's center. | \((-g, -f)\) |
| Radius Squared (from General Eq) | Square of the circle's radius. | \(r^2 = g^2 + f^2 - c\) |
| Point on Circle | Any point \((x, y)\) that satisfies the circle's equation. | \(x^2 + y^2 + 2gx + 2fy + c = 0\) holds true. |
Additional Information: Geometric Interpretation
Geometrically, the center of the circle is equidistant from all points on the circle. The two given points (3, -2) and (-2, 0) lie on the circle, so the distance from the center (-g, -f) to (3, -2) must be equal to the distance from the center (-g, -f) to (-2, 0). This distance is the radius of the circle.
Also, the center of the circle lies on the given line 2x – y – 3 = 0. This linear equation represents a straight line that contains the center. The intersection of the locus of points equidistant from (3, -2) and (-2, 0) (which is the perpendicular bisector of the segment connecting these two points) and the line 2x – y – 3 = 0 would give the center of the circle. Our algebraic method using the general equation and substituting points and line condition effectively finds this intersection point (the center) and the radius (implicitly through 'c').
Important Questions from Equation of Circle
The equation x2 + y2 + 2gx + 2fy + c = 0 always represents a circle whose centre is (-g, -f) and radius is \(\sqrt {{g^2} + {f^2} - c} \). If g2 + f2 = c, then in this case, the circle is called as:
The equation of circle with centre (1, -2) and radius 4 cm is:
The intercept on the line y = x by the circle x 2+ y 2- 2x = 0 is AB. Equation of circle with AB as diameter is
If the equation x 2+ y 2 - 4x - 4y + 4 = 0 represents a circle, then its radius is
Radius of the circle x 2+ y 2– 4x + 2y – 31 = 0 is
