Let the tangent to the circle C1 : x² + y² = 2 at the point M(-1, 1) intersect the circle C2 : (x -

Q: Why do we use the chord of contact formula instead of finding the intersection points $A$ and $B$ directly?

Finding the intersection points $A$ and $B$ by solving the quadratic equation would lead to messy, irrational coordinates. The chord of contact property allows us to relate the geometry of the tangent directly to the coordinates of $N$ without ever needing to find $A$ and $B$ explicitly. It is the elegant path!

Q: How can I be sure that the line $x - y + 2 = 0$ is the chord of contact for $N$?

The problem states that $A$ and $B$ are the points of tangency from $N$ to $C_2$. By definition, the line segment connecting the points of tangency from an external point is the chord of contact. Since $A$ and $B$ lie on the line $x - y + 2 = 0$, that line must be the chord of contact.

Q: Is there a simpler way to find the area of $\triangle ANB$?

Once you have the coordinates of $N$ and the equation of the line $AB$, the area is simply $\frac{1}{2} \times \text{base} \times \text{height}$. The base is the length of the chord $AB$, and the height is the perpendicular distance from $N$ to the line $AB$. This is the most direct and reliable method.

Question

Let the tangent to the circle $C_1 : x^2 + y^2 = 2$ at the point $M(-1, 1)$ intersect the circle $C_2 : (x - 3)^2 + (y - 2)^2 = 5$ at two distinct points $A$ and $B$. If the tangents to $C_2$ at the points $A$ and $B$ intersect at $N$, then the area of the triangle $ANB$ is equal to :

Options:

(A) $\frac{1}{2}$

(B) $\frac{2}{3}$

(C) $\frac{1}{6}$

(D) $\frac{5}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with circle $C_1$ centered at the origin with radius $\sqrt{2}$, defined by the equation $x^2 + y^2 = 2$. A point $M(-1, 1)$ lies on its circumference.

To find the tangent at $M$, we apply the $T=0$ transformation, replacing $x^2$ with $x x_1$ and $y^2$ with $y y_1$. Substituting $M(-1, 1)$, we obtain:
$$-x + y = 2 \quad \Rightarrow \quad x - y + 2 = 0$$
This line serves as our primary geometric constraint.

### The Intersection

We introduce circle $C_2$, defined by the equation:
$$(x - 3)^2 + (y - 2)^2 = 5$$
This circle is centered at $(3, 2)$ with a radius $R = \sqrt{5}$. The line $x - y + 2 = 0$ intersects $C_2$ to form a chord $AB$.

We are given that the tangents to $C_2$ at points $A$ and $B$ intersect at a point $N(h, k)$. Consequently, the line $AB$ is the **chord of contact** for point $N$ with respect to $C_2$.

### The Bridge of Logic

The equation for the chord of contact from $N(h, k)$ to the circle $(x - 3)^2 + (y - 2)^2 = 5$ is given by:
$$(h - 3)(x - 3) + (k - 2)(y - 2) = 5$$
Expanding this expression, we get:
$$(h - 3)x + (k - 2)y - (3h + 2k - 18) = 0$$

Since this represents the same line as $x - y + 2 = 0$, the coefficients must be proportional:
$$\frac{h - 3}{1} = \frac{k - 2}{-1} = \frac{-(3h + 2k - 18)}{2}$$

Solving these ratios, we obtain the system:
1. $h + k = 5$
2. $5h + 2k = 14$

Solving this system yields the coordinates of the intersection point:
$$N = \left( \frac{4}{3}, \frac{11}{3} ight)$$

### Final Calculation

The area of $	riangle ANB$ is calculated using $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. First, we find the length of the chord $AB$. The distance $d$ from the center $(3, 2)$ to the line $x - y + 2 = 0$ is:
$$d = \frac{|3 - 2 + 2|}{\sqrt{1^2 + (-1)^2}} = \frac{3}{\sqrt{2}}$$

The chord length $AB$ is given by $2\sqrt{R^2 - d^2}$:
$$AB = 2\sqrt{5 - \frac{9}{2}} = 2\sqrt{\frac{1}{2}} = \sqrt{2}$$

Next, the height $p$ is the perpendicular distance from $N(\frac{4}{3}, \frac{11}{3})$ to the line $x - y + 2 = 0$:
$$p = \frac{|\frac{4}{3} - \frac{11}{3} + 2|}{\sqrt{1^2 + (-1)^2}} = \frac{|-\frac{7}{3} + \frac{6}{3}|}{\sqrt{2}} = \frac{1}{3\sqrt{2}}$$

Finally, the area of $	riangle ANB$ is:
$$	ext{Area} = \frac{1}{2} 	imes \sqrt{2} 	imes \frac{1}{3\sqrt{2}} = \mathbf{\frac{1}{6}}$$

Select Answer:

Visualized Solution

Visualizing Circle $C_{1}$ and Point $M$

Finding the Tangent Equation at $M$

Introducing Circle $C_{2}$

The Concept of Chord of Contact

Equation of Chord of Contact from $N$

Comparing the Two Equations

Solving for Point $N$

Calculating Chord Length $A B$

Calculating Height of $△ A N B$

Final Area Calculation

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

The Intersection

The Bridge of Logic

Final Calculation

Similar Questions

If the tangents at the points $P$ and $Q$ on the circle $x^{2} + y^{2} - 2 x + y = 5$ meet at the point $R (\frac{9}{4}, 2)$ , then the area of the triangle $P QR$ is

Let $B$ be the centre of the circle $x^{2} + y^{2} - 2 x + 4 y + 1 = 0$ . Let the tangents at two points $P$ and $Q$ on the circle intersect at the point $A (3, 1)$ . Then $8 \cdot (\frac{area Δ A P Q}{area Δ B P Q})$ is equal to .

The area of the triangle formed by the tangents from the point $(4, 3)$ to the the circle $x^{2} + y^{2} = 9$ and the line joining their points of contact is .........

A circle with centre $(2, 3)$ and radius 4 intersects the line $x + y = 3$ at the points $P$ and $Q$ . If the tangents at $P$ and $Q$ intersect at the point $S (α, β)$ , then $4 α - 7 β$ is equal to

If the tangent at the point $P$ on the circle $x^{2} + y^{2} + 6 x + 6 y = 2$ meets a straight line $5 x - 2 y + 6 = 0$ at a point $Q$ on the y-axis, then the length of $P Q$ is

Let the tangents drawn from the origin to the circle, $x^{2} + y^{2} - 8 x - 4 y + 16 = 0$ touch it at the points $A$ and $B$ . The $(A B)^{2}$ is equal to :

Let the tangents drawn from the origin to the circle, $x^{2} + y^{2} - 8 x - 4 y + 16 = 0$ touch it at the point $A$ and $B$ . The $(A B)^{2}$ is equal to:

Comprehension Passage

Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

Let $x^{2} + y^{2} - 4 x - 2 y - 11 = 0$ be a circle. A pair of tangents from the point $(4, 5)$ with a pair of radii form a quadrilateral of area .........

Select Answer:

Visualized Solution

Visualizing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​ and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Finding the Tangent Equation at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Introducing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​