Let B be the centre of the circle x² + y² - 2x + 4y + 1 = 0. Let the tangents at two points P and Q

Q: Wait, why is $\Delta APB$ a right-angled triangle? I thought the tangent was just a line.

That is a fantastic question! Remember the fundamental property of circles: the radius drawn to the point of tangency is always perpendicular to the tangent line. Since $BP$ is the radius and $AP$ is the tangent, the angle $\angle APB$ must be $90^\circ$. This geometric fact is the key that unlocks the entire problem!

Q: Why does the area ratio $\frac{\text{Area } \Delta APQ}{\text{Area } \Delta BPQ}$ simplify to just $\frac{AR}{RB}$?

Great observation! Both triangles, $\Delta APQ$ and $\Delta BPQ$, share the same base, $PQ$. The area of any triangle is $\frac{1}{2} \cdot \text{base} \cdot \text{height}$. When you take the ratio of these two areas, the $\frac{1}{2}$ and the base $PQ$ cancel out completely, leaving you with the ratio of their respective heights, $AR$ and $RB$. It is a beautiful simplification!

Q: Can I just find the coordinates of $P$ and $Q$ to solve this?

You absolutely could, but it would be a long and winding road! You would need to find the equation of the chord of contact, solve it with the circle equation, find the coordinates of $P$ and $Q$, and then use the area formula. While that works, using the geometric properties of the right triangle $\Delta APB$ is much faster and less prone to calculation errors. In JEE, elegance is often rewarded with speed!

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Every great geometric story begins with understanding the protagonist. We are given the circle equation: 
$$x^2 + y^2 - 2x + 4y + 1 = 0$$

To understand its nature, we must find its center and radius. By comparing this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify $g = -1$, $f = 2$, and $c = 1$.

The center $B$ is at $(-g, -f)$, which gives us $B(1, -2)$. The radius $r$ is calculated as:
$$r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + 2^2 - 1} = \sqrt{1 + 4 - 1} = \sqrt{4} = 2$$

Now, we know our circle intimately: centered at $(1, -2)$ with a radius of $2$.

### The Power of the Tangent

We are given an external point $A(3, 1)$. From this point, two tangents are drawn to the circle, touching it at $P$ and $Q$.

The length of these tangents is a classic calculation. Using the formula $L = \sqrt{S_1}$, where $S_1$ is the power of the point $A$ with respect to the circle, we substitute $A(3, 1)$ into the circle equation:
$$S_1 = 3^2 + 1^2 - 2(3) + 4(1) + 1 = 9 + 1 - 6 + 4 + 1 = 9$$

Thus, the length of the tangent $AP = AQ = \sqrt{9} = 3$.

### The Hidden Right Triangle

Now, let us connect the center $B$ to the external point $A$. The distance $AB$ is:
$$AB = \sqrt{(3 - 1)^2 + (1 - (-2))^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$$

Here is where the magic happens. Consider the triangle $\Delta APB$. Because the radius $BP$ is perpendicular to the tangent $AP$, $\angle APB = 90^\circ$.

If we draw the chord of contact $PQ$, it intersects $AB$ at a point $R$. Since $AB$ is the perpendicular bisector of the chord $PQ$, $PR$ is the altitude to the hypotenuse $AB$.

Using the properties of right-angled triangles, we know that $AP^2 = AR \cdot AB$ and $BP^2 = RB \cdot AB$. We can now find the lengths of the segments $AR$ and $RB$:

$$AR = \frac{AP^2}{AB} = \frac{3^2}{\sqrt{13}} = \frac{9}{\sqrt{13}}$$

$$RB = \frac{BP^2}{AB} = \frac{2^2}{\sqrt{13}} = \frac{4}{\sqrt{13}}$$

### The Elegant Cancellation

We need the ratio of the areas of $\Delta APQ$ and $\Delta BPQ$. The area of $\Delta APQ$ is $\frac{1}{2} \cdot PQ \cdot AR$, and the area of $\Delta BPQ$ is $\frac{1}{2} \cdot PQ \cdot RB$.

When we divide these, the $\frac{1}{2}$ and the $PQ$ vanish into thin air! We are left with the ratio $\frac{AR}{RB}$.

Substituting our values:
$$\frac{AR}{RB} = \frac{9/\sqrt{13}}{4/\sqrt{13}} = \frac{9}{4}$$

### Final Calculation

The question asks for $8 \cdot \left( \frac{	ext{Area } \Delta APQ}{	ext{Area } \Delta BPQ} ight)$.

With our ratio in hand, the calculation is simple:
$$8 \cdot \frac{9}{4} = 2 \cdot 9 = 18$$

We have arrived at the answer, **18**. Whenever you face a daunting problem, look for the right triangles, look for the perpendiculars, and trust in the elegance of the math.

Animated Solution for Mathematics - Circles: 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 .

Enter Numerical Value:

Visualized Solution

Visualizing the Problem Setup

Finding Center $B$ and Radius $r$

Length of Tangent $A P$

Distance Between $A$ and $B$

Geometry of the Triangles

Simplifying the Area Ratio

Calculating Segments $A R$ and $R B$

Final Ratio and Result

The Sigma Insight: Length of Tangent and Chord of Contact

Analyzing the Setup

The Power of the Tangent

The Hidden Right Triangle

The Elegant Cancellation

Final Calculation

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