Chord of Contact and Radical Axis

Chord of Contact and Radical Axis

Hello JEE aspirants! This lesson dives into some crucial concepts related to circles that can significantly boost your score in Coordinate Geometry. We'll explore chords of contact, radical axes, and families of circles. These topics are frequently tested in JEE Main, often in combination with other concepts, so mastering them is key!

1. Chord of Contact from an External Point

Imagine a circle and a point outside it. You can draw two tangents from that external point to the circle. The chord joining the points where these tangents touch the circle is called the chord of contact. It's a fundamental geometric concept with a neat algebraic representation.

Geometric Interpretation: The chord of contact is the locus of points that are the feet of the perpendiculars drawn from the external point to the tangent lines of the circle. This might sound complicated, but visualize it! The chord of contact is intimately tied to the tangent properties of the circle.

Formula:

Here's what $T = 0$ means. If your circle's equation is $S = x^2 + y^2 + 2gx + 2fy + c = 0$, then $T$ is obtained by replacing:

• $x^2$ with $xx_1$
• $y^2$ with $yy_1$
• $x$ with $\frac{x + x_1}{2}$
• $y$ with $\frac{y + y_1}{2}$

So, the equation of the chord of contact becomes: $T = xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$

Derivation/Explanation: Let the equation of the circle be $S = x^2 + y^2 + 2gx + 2fy + c = 0$. Let the external point be $P(x_1, y_1)$. Let the points of tangency be $A(x_a, y_a)$ and $B(x_b, y_b)$. The equations of the tangents at A and B are: $T_A = 0$ and $T_B = 0$, respectively. Since P lies on both tangents, $T_A(x_1, y_1) = 0$ and $T_B(x_1, y_1) = 0$. This means $A$ and $B$ both satisfy the equation $T(x_1, y_1) = 0$. Therefore, $T = 0$ represents the equation of the chord $AB$, which is the chord of contact.

Example: Find the equation of the chord of contact of the circle $x^2 + y^2 = 4$ from the point $(3, 2)$. Here, $x_1 = 3$, $y_1 = 2$, $g = 0$, $f = 0$, and $c = -4$. Therefore, $T = 3x + 2y - 4 = 0$.

2. Chord with a Given Midpoint

Now, consider a chord of the circle where you know the coordinates of its midpoint. How do you find the equation of that chord? This scenario also has a very elegant solution.

Geometric Interpretation: The chord with a given midpoint is perpendicular to the line joining the center of the circle and the midpoint. Use this fact for alternative solutions!

Formula:

Here, $T$ is the same as in the chord of contact formula. $S_1$ means you substitute $(x_1, y_1)$ into the circle's equation $S$. So, $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$. The equation of the chord with midpoint $(x_1, y_1)$ is then: $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$.

Derivation/Explanation: The slope of the line joining the center $(-g, -f)$ and the midpoint $(x_1, y_1)$ is $m_1 = \frac{y_1 + f}{x_1 + g}$. The slope of the chord is $m_2 = -\frac{x_1 + g}{y_1 + f}$ (since they are perpendicular). The equation of the chord is $y - y_1 = m_2(x - x_1)$. After simplification, and using the fact that $(x_1, y_1)$ lies on the chord, we arrive at $T = S_1$. This derivation reinforces the geometric interpretation!

Example: Find the equation of the chord of the circle $x^2 + y^2 - 4x + 6y - 12 = 0$ with midpoint $(1, 1)$. Here, $x_1 = 1$, $y_1 = 1$, $g = -2$, $f = 3$, and $c = -12$. So, $S_1 = 1 + 1 - 4 + 6 - 12 = -8$. Thus, $T = S_1$ gives $x + y - 2(x + 1) + 3(y + 1) - 12 = -8$, which simplifies to $-x + 4y - 3 = 0$ or $x - 4y + 3 = 0$.

3. Radical Axis of Two Circles

Consider two circles. The radical axis is the locus of a point such that the lengths of the tangents from it to the two circles are equal.

Geometric Interpretation: The radical axis is a straight line. It's perpendicular to the line joining the centers of the two circles. If the circles intersect, the radical axis is simply the common chord.

Formula:

If $S_1 = x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $S_2 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$, then the equation of the radical axis is: $2(g_1 - g_2)x + 2(f_1 - f_2)y + (c_1 - c_2) = 0$.

Derivation/Explanation: Let $P(x, y)$ be a point on the radical axis. Let the lengths of the tangents from $P$ to the circles $S_1 = 0$ and $S_2 = 0$ be $L_1$ and $L_2$ respectively. Then $L_1^2 = S_1(x, y)$ and $L_2^2 = S_2(x, y)$. Since $L_1 = L_2$, we have $S_1(x, y) = S_2(x, y)$, which gives $S_1 - S_2 = 0$.

Example: Find the equation of the radical axis of the circles $x^2 + y^2 + 2x - 4y - 4 = 0$ and $x^2 + y^2 - 6x + 2y + 6 = 0$. Here, $S_1 - S_2 = (2 - (-6))x + (-4 - 2)y + (-4 - 6) = 0$, which simplifies to $8x - 6y - 10 = 0$ or $4x - 3y - 5 = 0$.

4. Radical Center of Three Circles

Given three circles, the radical center is the point of concurrency of the three radical axes taken pairwise.

Geometric Interpretation: The radical center is the point from which the lengths of the tangents to the three circles are equal. If the centers of the three circles are non-collinear, then the radical center is a unique point.

Finding the Radical Center: Find the equations of the radical axes of circles 1 & 2 ($S_1 - S_2 = 0$) and circles 2 & 3 ($S_2 - S_3 = 0$). Solve these two equations simultaneously to find the coordinates of the radical center.

Example: Find the radical center of the circles $x^2 + y^2 = 4$, $x^2 + y^2 + 2x + 4y + 1 = 0$, and $x^2 + y^2 - 4x + 2y - 6 = 0$. Radical axis 1 (circles 1 & 2): $2x + 4y + 5 = 0$ Radical axis 2 (circles 2 & 3): $6x + 2y + 7 = 0$ Solving these two equations gives the radical center. Multiply the first equation by 3 to get $6x + 12y + 15 = 0$. Subtracting the second equation from this gives $10y + 8 = 0$, so $y = -\frac{4}{5}$. Substituting this into $2x + 4y + 5 = 0$ gives $2x - \frac{16}{5} + 5 = 0$, so $2x = -\frac{9}{5}$ and $x = -\frac{9}{10}$. The radical center is therefore $\left(-\frac{9}{10}, -\frac{4}{5}\right)$.

5. Family of Circles Through Intersection

If you have two circles, you can define a family of circles passing through their points of intersection.

Geometric Interpretation: This represents an infinite number of circles sharing the same common chord (the radical axis) as the original two circles.

Formula:

Where $\lambda$ is a parameter. By varying $\lambda$, you get different circles passing through the intersection of $S_1 = 0$ and $S_2 = 0$.

Derivation/Explanation: Let $S_1 = 0$ and $S_2 = 0$ represent the equations of the two circles. At the points of intersection of the two circles, both $S_1 = 0$ and $S_2 = 0$ are satisfied. Therefore, $S_1 + \lambda S_2 = 0$ is also satisfied at these points, regardless of the value of $\lambda$. This means that $S_1 + \lambda S_2 = 0$ represents a curve passing through the points of intersection of $S_1 = 0$ and $S_2 = 0$. Since $S_1$ and $S_2$ are both quadratic equations of a circle, $S_1 + \lambda S_2 = 0$ is also a quadratic equation, and it represents a circle (unless $\lambda = -1$, in which case it represents the common chord).

Example: Find the equation of the circle passing through the intersection of the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 2x + 4y - 4 = 0$, and also passing through the point $(1, 0)$. The family of circles is given by $x^2 + y^2 - 4 + \lambda(x^2 + y^2 - 2x + 4y - 4) = 0$. Since the circle passes through $(1, 0)$, we have $1 - 4 + \lambda(1 - 2 - 4) = 0$, so $-3 - 5\lambda = 0$, giving $\lambda = -\frac{3}{5}$. Substituting this value into the family equation, we get $x^2 + y^2 - 4 - \frac{3}{5}(x^2 + y^2 - 2x + 4y - 4) = 0$, which simplifies to $2x^2 + 2y^2 + \frac{6}{5}x - \frac{12}{5}y - \frac{8}{5} = 0$, or $10x^2 + 10y^2 + 6x - 12y - 8 = 0$, i.e., $5x^2 + 5y^2 + 3x - 6y - 4 = 0$.

That concludes our exploration of chord of contact and radical axis. Practice lots of problems to become comfortable with these concepts. All the best for your JEE preparation!