Equation of a Chord with known midpoint

Chord of a Circle with Known Midpoint

Let a circle be given by the equation

\[ x^2 + y^2 = a^2, \]

and suppose we are told that a chord of this circle has midpoint \( P(x_1, y_1) \). We want to find the equation of this chord.

By the symmetry of the circle, and its fundamental geometry, we proceed as follows:

Join the center \( O(0, 0) \) of the circle to the midpoint \( P(x_1, y_1) \) of the chord. Then, from classical geometry of the circle, this line \( OP \) is perpendicular to the chord. Hence, the slope of the chord is the negative reciprocal of the slope of \( OP \).

The slope of \( OP \) is:

\[ \frac{y_1}{x_1} \quad (\text{assuming } x_1 \neq 0), \]

so the slope of the chord is:

\[ - \frac{x_1}{y_1}. \]

We now have the slope of the chord and a point \( (x_1, y_1) \) through which it passes. Using point-slope form, the equation of the chord is:

\[ \frac{y - y_1}{x - x_1} = -\frac{x_1}{y_1}. \]

Cross-multiplying:

\[ y y_1 - y_1^2 = -x x_1 + x_1^2. \]

Rewriting:

\[ x x_1 + y y_1 = x_1^2 + y_1^2. \]

Now subtract \( a^2 \) from both sides:

\[ x x_1 + y y_1 - a^2 = x_1^2 + y_1^2 - a^2. \]

On the left-hand side, the expression \( x x_1 + y y_1 - a^2 \) is the tangent form \( T \) for the circle \( x^2 + y^2 = a^2 \). On the right-hand side, the expression \( x_1^2 + y_1^2 - a^2 \) is the value of the circle at \( (x_1, y_1) \), denoted \( S_1 \).

Thus, the required equation of the chord is:

\[ \boxed{T = S_1}, \quad \text{or} \quad \boxed{ x x_1 + y y_1 - a^2 = x_1^2 + y_1^2 - a^2 }. \]

This form generalizes beautifully. It says:

The equation of a chord of a circle, given its midpoint \( (x_1, y_1) \), is \( T = S_1 \).

In case of a general circle \(x^2+y^2+2gx+2fy+c=0\) this equation of the chord is \(xx_1+yy_1+g(x+x_1)+f(y+y_1)+c = x_1^2+y_1^2+2gx_1+2fy_1+c=0\).

Example

Given the circle

\[ x^2 + y^2 + 4x - 3y - 3 = 0, \]

and the midpoint of a chord is \( (-3,\ 1) \), find the equation of the chord.

Solution:

We use the identity \( T = S_1 \), where:

• \( T \) is obtained by replacing \( x^2 \to x(-3),\ y^2 \to y(1),\ x \to \frac{x - 3}{2},\ y \to \frac{y + 1}{2} \) in the left-hand side of the circle equation,
• \( S_1 \) is the value of the circle's expression at \( x = -3,\ y = 1 \).

Thus,

\[ T = -3x + y + 4 \cdot \frac{x - 3}{2} - 3 \cdot \frac{y + 1}{2} - 3,\quad S_1 = (-3)^2 + 1^2 + 4(-3) - 3(1) - 3 = -8. \]

Hence,

\[ T = S_1 \implies -3x + y + 2(x - 3) - \tfrac{3}{2}(y + 1) - 3 = -8. \]

Simplifying directly:

\[ - x - \tfrac{1}{2}y - \tfrac{21}{2} = -8 \implies 2x + y = -5. \]

\[ \boxed{2x + y + 5 =0} \]

is the required equation of the chord.

Example

Find the locus of the midpoint of a chord of the circle

\[ x^2 + y^2 = 7 \]

if the chord always passes through the fixed point \( (-1,\ 2) \).

Solution:

Let the midpoint of such a chord be \( (h,\ k) \). The equation of the chord with this midpoint is given by the identity:

\[ T = S_1. \]

For the circle \( x^2 + y^2 = 7 \), this gives:

\[ xh + yk - 7 = h^2 + k^2 - 7 \implies xh + yk = h^2 + k^2. \tag{1} \]

This is the equation of the chord with midpoint \( (h,\ k) \). But it is given that this chord always passes through the point \( (-1,\ 2) \), so substitute \( x = -1, y = 2 \) into (1):

\[ (-1)h + 2k = h^2 + k^2 \implies -h + 2k = h^2 + k^2. \tag{2} \]

Now, replace \( h \to x \), \( k \to y \) to obtain the locus of the midpoint \( (h,\ k) \):

\[ x^2 + y^2 + x - 2y = 0. \]

Hence, the required locus is:

\[ \boxed{x^2 + y^2 + x - 2y = 0}. \]