Two tangents are drawn from the point P(-1, 1) to the circle x² + y² - 2x - 6y + 6 = 0. If these tan

Q: Wait, why is the quadrilateral $PACB$ a square? I thought it was just a kite.

That is a fantastic observation! In any external point tangent problem, $PACB$ is indeed a kite because $PA=PB$ and $CA=CB$. However, in this specific case, the length of the tangent $L$ is $2$, and the radius $r$ is also $2$. Because the adjacent sides $PA$ and $CA$ are both equal to $2$, the kite is elevated to the status of a square. It is a special case where the geometry simplifies beautifully!

Q: How do we know $D$ is $(3, 3)$ and not $B(-1, 3)$?

Great catch! The problem states that $D$ is a point on the circle such that $AD = AB$. We found that $\angle ACD = 90^\circ$. Since $\vec{CA}$ is vertical, $\vec{CD}$ must be horizontal. Moving $2$ units horizontally from $C(1, 3)$ gives us two points: $(1-2, 3) = (-1, 3)$ and $(1+2, 3) = (3, 3)$. The point $(-1, 3)$ is already point $B$. Since $D$ must be a distinct point on the circle, we must choose $(3, 3)$.

Q: Is there a faster way to find the area of $\triangle ABD$ than using base and height?

You could use the determinant formula for the area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$, which is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. However, in this problem, the coordinates are so clean that the base-height method is much faster and less prone to calculation errors. Always look for horizontal or vertical lines first!

Question

Two tangents are drawn from the point $P(-1, 1)$ to the circle $x^2 + y^2 - 2x - 6y + 6 = 0$. If these tangents touch the circle at points $A$ and $B$, and if $D$ is a point on the circle such that length of the segments $AB$ and $AD$ are equal, then the area of the triangle $ABD$ is equal to:

Options:

(A) 2

(B) $(3\sqrt{2} + 2)$

(C) 4

(D) $3(\sqrt{2} - 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given equation is $x^2 + y^2 - 2x - 6y + 6 = 0$. To reveal the geometry of this circle, we complete the square for the $x$ and $y$ terms:

$$(x^2 - 2x + 1) + (y^2 - 6y + 9) = -6 + 1 + 9$$

This simplifies to the standard form:

$$(x - 1)^2 + (y - 3)^2 = 4$$

From this, we identify the center $C$ at $(1, 3)$ and the radius $r = 2$.

### The Tangent Mystery

Consider the point $P(-1, 1)$ located outside the circle. We draw two tangents from $P$ to the circle, touching at points $A$ and $B$. The length $L$ of these tangents is given by the power of the point formula, $L = \sqrt{S_1}$.

Substituting $P(-1, 1)$ into the circle equation:

$$S_1 = (-1)^2 + (1)^2 - 2(-1) - 6(1) + 6 = 1 + 1 + 2 - 6 + 6 = 4$$

Thus, the tangent length is $L = \sqrt{4} = 2$. Since the radius $r$ is also $2$, we have $PA = PB = CA = CB = 2$.

### The Elegance of the Square

In the quadrilateral $PACB$, the angles at the points of contact are right angles, meaning $\angle PAC = 90^\circ$ and $\angle PBC = 90^\circ$. Because all sides are equal to $2$ and the adjacent angles are $90^\circ$, $PACB$ is a square.

Given the orientation of the center $C(1, 3)$ and the tangent lengths, we determine the coordinates of the contact points. By moving $2$ units from the center, we find:

$$A = (1, 1)$$
$$B = (-1, 3)$$

### The Hunt for Point $D$

We seek a point $D$ on the circle such that $AD = AB$. First, we calculate the length of the chord $AB$:

$$AB = \sqrt{(1 - (-1))^2 + (1 - 3)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$$

In $	riangle ACD$, we have $AC = 2$, $CD = 2$, and $AD = 2\sqrt{2}$. Since $AC^2 + CD^2 = 2^2 + 2^2 = 8 = AD^2$, $	riangle ACD$ is a right-angled triangle with $\angle ACD = 90^\circ$.

Since $\vec{CA}$ is a vertical vector, $\vec{CD}$ must be horizontal. Moving $2$ units horizontally from $C(1, 3)$ yields points $(-1, 3)$ and $(3, 3)$. Since $(-1, 3)$ is point $B$, we conclude $D = (3, 3)$.

### The Final Area

We now calculate the area of $	riangle ABD$ with vertices $A(1, 1)$, $B(-1, 3)$, and $D(3, 3)$. Since $B$ and $D$ share the same $y$-coordinate, the base $BD$ is horizontal:

$$	ext{Base } BD = |3 - (-1)| = 4$$

The height of the triangle is the vertical distance from $A(1, 1)$ to the line $y = 3$:

$$	ext{Height} = |3 - 1| = 2$$

The area of the triangle is:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 2 = 4$$

The final area of $	riangle ABD$ is **4**.

Select Answer:

Visualized Solution

Standardizing the Circle Equation

Calculating Tangent Length $L$

Geometric Deduction: Square $P A C B$

Finding Coordinates of $A$ and $B$

Calculating Chord Length $A B$

Geometric Deduction for Point $D$

Locating Point $D$

Final Area Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent Mystery

The Elegance of the Square

The Hunt for Point $D$

The Final Area

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Visualized Solution

Standardizing the Circle Equation

Calculating Tangent Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Geometric Deduction: Square .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PACB

Finding Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Calculating Chord Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Geometric Deduction for Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Locating Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D