A circle is drawn on the chord of a circle x 2+ y 2= a 2as diameter. The chord lies on the line x + y = a. What is the equation of the circle?
x 2+ y 2– ax – ay = 0
Understanding the Problem: Circle on a Chord as Diameter
The question asks for the equation of a new circle. This new circle has a specific property: its diameter is the chord formed by the intersection of an existing circle \(\small x^2 + y^2 = a^2\) and a line \(\small x + y = a\).
Let's break down the given information:
The original circle is \(\small S_1: x^2 + y^2 = a^2\). We can write this as \(\small S_1: x^2 + y^2 - a^2 = 0\).
The line is \(\small L: x + y = a\). We can write this as \(\small L: x + y - a = 0\).
The new circle passes through the intersection points of the original circle and the line. The equation of any circle passing through the intersection points of a circle \(\small S=0\) and a line \(\small L=0\) is given by the general form:
\(\small S + \lambda L = 0\)
where \(\small \lambda\) is a constant.
Applying the Family of Circles Formula
Using the given equations, the equation of the required circle is:
\(\small (x^2 + y^2 - a^2) + \lambda (x + y - a) = 0\)
This can be rearranged as:
\(\small x^2 + y^2 + \lambda x + \lambda y - a^2 - \lambda a = 0\)
This equation represents a family of circles passing through the intersection points of the original circle and the line.
Using the Diameter Condition to Find Lambda
We are given that the chord \(\small x + y = a\) is the diameter of the new circle. A key property is that the center of a circle always lies on its diameter.
Let's find the center of the new circle \(\small x^2 + y^2 + \lambda x + \lambda y - a^2 - \lambda a = 0\).
Comparing this to the standard form of a circle equation \(\small x^2 + y^2 + 2gx + 2fy + c = 0\), the center is at \(\small (-g, -f)\).
From our equation, we have:
\(\small 2g = \lambda \Rightarrow g = \frac{\lambda}{2}\)
\(\small 2f = \lambda \Rightarrow f = \frac{\lambda}{2}\)
So, the center of the new circle is \(\small \left(-\frac{\lambda}{2}, -\frac{\lambda}{2}\right)\).
Since the chord \(\small x + y = a\) is the diameter, the center \(\small \left(-\frac{\lambda}{2}, -\frac{\lambda}{2}\right)\) must lie on this line.
Substitute the coordinates of the center into the line equation \(\small x + y = a\):
\(\small -\frac{\lambda}{2} + -\frac{\lambda}{2} = a\)
\(\small -\lambda = a\)
\(\small \lambda = -a\)
Substituting Lambda to Find the Circle Equation
Now we substitute the value of \(\small \lambda = -a\) back into the equation of the family of circles:
\(\small (x^2 + y^2 - a^2) + (-a) (x + y - a) = 0\)
\(\small x^2 + y^2 - a^2 - ax - ay + a^2 = 0\)
The \(\small -a^2\) and \(\small +a^2\) terms cancel out.
The equation of the required circle is:
\(\small x^2 + y^2 - ax - ay = 0\)
Comparing with Options
Let's compare our derived equation with the given options:
| Option | Equation |
|---|---|
| 1 | \(\small x^2 + y^2 – ax – ay + a^2 = 0\) |
| 2 | \(\small x^2 + y^2 – ax – ay = 0\) |
| 3 | \(\small x^2 + y^2 + ax + ay = 0\) |
| 4 | \(\small x^2 + y^2 + ax + ay – 2a^2 = 0\) |
Our derived equation \(\small x^2 + y^2 - ax - ay = 0\) matches Option 2.
Revision Table: Key Concepts for Circle Equations
| Concept | Description | Formula/Method |
|---|---|---|
| Equation of a Circle | Standard form centered at origin (0,0) | \(\small x^2 + y^2 = r^2\) |
| Equation of a Circle | General form | \(\small x^2 + y^2 + 2gx + 2fy + c = 0\) (Center: \(\small (-g, -f)\), Radius: \(\small \sqrt{g^2+f^2-c}\)) |
| Family of Circles | Passing through intersection of circle \(\small S=0\) and line \(\small L=0\) | \(\small S + \lambda L = 0\) |
| Family of Circles | Passing through intersection of circles \(\small S_1=0\) and \(\small S_2=0\) | \(\small S_1 + \lambda S_2 = 0\) (\(\small \lambda \neq -1\) for a circle) |
| Diameter Property | Center of circle lies on its diameter | If line \(\small Ax+By+C=0\) is diameter, center \(\small (-g, -f)\) satisfies \(\small A(-g)+B(-f)+C=0\) |
Additional Information: Understanding S + λL = 0
The equation \(\small S + \lambda L = 0\) is a powerful tool in coordinate geometry for finding equations of curves passing through the intersection points of two given curves. In this case, the first curve is the original circle (\(\small S=0\)) and the second "curve" is the line (\(\small L=0\)).
Any point that lies on both the circle \(\small S=0\) and the line \(\small L=0\) satisfies both equations simultaneously.
For such a point, \(\small S=0\) and \(\small L=0\). Therefore, substituting these values into \(\small S + \lambda L = 0\) gives \(\small 0 + \lambda(0) = 0\), which is always true. This shows that the equation \(\small S + \lambda L = 0\) always passes through the intersection points, regardless of the value of \(\small \lambda\).
By varying \(\small \lambda\), we get different circles (or a line, if \(\small \lambda\) makes the \(\small x^2\) and \(\small y^2\) terms cancel, which doesn't happen here as there are no \(\small x^2\) or \(\small y^2\) terms in \(\small L\)).
The specific value of \(\small \lambda\) is determined by an additional condition given in the problem. In this case, the condition is that the chord is the diameter, meaning the center lies on the line \(\small L=0\). Other conditions might be that the circle passes through another given point, or has a specific radius, etc.
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
