If the tangents drawn at the point O(0, 0) and P(1 + √5, 2) on the circle x² + y² - 2x - 4y = 0 inte

Q: Wait, why do we use the $T=0$ method instead of finding the slope of the radius and taking the negative reciprocal?

Both methods are perfectly valid! The $T=0$ method is just a faster, more algebraic shortcut that works for any conic section. Finding the radius slope and taking the negative reciprocal is the geometric approach, which is excellent for building intuition. If you prefer the geometric way, go for it—it will lead to the exact same result!

Q: Why is the tangent at $P$ a vertical line? Did I make a mistake?

You did not make a mistake! If you calculate the slope of the radius from the center $(1, 2)$ to $P(1 + \sqrt{5}, 2)$, you will find it is zero (a horizontal line). Since the tangent is perpendicular to the radius, it must be a vertical line. It is a beautiful consequence of the coordinates provided.

Q: Is there a faster way to find the area of triangle $OPQ$ without using the determinant formula?

Absolutely. Since $PQ$ is a vertical line, you can treat it as the base of the triangle. The length of the base is simply the difference in the y-coordinates of $P$ and $Q$. The height is the horizontal distance from the origin $O$ to the vertical line $PQ$. Using $\frac{1}{2} \times \text{base} \times \text{height}$ is much faster than the coordinate determinant formula here.

Question

If the tangents drawn at the point $O(0, 0)$ and $P(1 + \sqrt{5}, 2)$ on the circle $x^2 + y^2 - 2x - 4y = 0$ intersect at the point $Q$, then the area of the triangle $OPQ$ is equal to

Options:

(A) $\frac{3 + \sqrt{5}}{2}$

(B) $\frac{4 + 2\sqrt{5}}{2}$

(C) $\frac{5 + 3\sqrt{5}}{2}$

(D) $\frac{7 + 3\sqrt{5}}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given circle is defined by the equation $x^2 + y^2 - 2x - 4y = 0$. We are working with two points on this circle: the origin $O(0, 0)$ and the point $P(1 + \sqrt{5}, 2)$.

Our objective is to determine the area of the triangle formed by these two points and the intersection point $Q$ of the tangents drawn to the circle at $O$ and $P$.

### The Power of $T=0$

To find the tangents at these points, we utilize the $T=0$ method. For a circle, the tangent at $(x_1, y_1)$ is found by replacing $x^2 	o xx_1$, $y^2 	o yy_1$, $x 	o \frac{x+x_1}{2}$, and $y 	o \frac{y+y_1}{2}$.

Applying this to the origin $O(0, 0)$:
$$x(0) + y(0) - (x + 0) - 2(y + 0) = 0$$
This simplifies to the line:
$$x + 2y = 0$$

For point $P(1 + \sqrt{5}, 2)$, the substitution yields:
$$x(1 + \sqrt{5}) + y(2) - (x + 1 + \sqrt{5}) - 2(y + 2) = 0$$

Expanding and simplifying the terms:
$$x + \sqrt{5}x + 2y - x - 1 - \sqrt{5} - 2y - 4 = 0$$
$$\sqrt{5}x - 5 - \sqrt{5} = 0$$
Dividing by $\sqrt{5}$, we obtain the vertical line:
$$x = 1 + \sqrt{5}$$

### The Intersection and the Triangle

To find the intersection point $Q$, we substitute $x = 1 + \sqrt{5}$ into the first tangent equation $x + 2y = 0$:
$$1 + \sqrt{5} + 2y = 0 \Rightarrow y = -\frac{1 + \sqrt{5}}{2}$$
Thus, the coordinates of $Q$ are $(1 + \sqrt{5}, -\frac{1 + \sqrt{5}}{2})$.

Consider the triangle $OPQ$. Since $P$ and $Q$ share the same $x$-coordinate, the segment $PQ$ is a vertical line segment.

The length of the base $PQ$ is the vertical distance:
$$b = |2 - (-\frac{1 + \sqrt{5}}{2})| = \frac{4 + 1 + \sqrt{5}}{2} = \frac{5 + \sqrt{5}}{2}$$

The height $h$ of the triangle is the horizontal distance from the origin $O(0, 0)$ to the vertical line $x = 1 + \sqrt{5}$:
$$h = 1 + \sqrt{5}$$

### Final Calculation

The area of the triangle is given by the formula $	ext{Area} = \frac{1}{2} 	imes b 	imes h$. Substituting our values:

$$	ext{Area} = \frac{1}{2} 	imes \left( \frac{5 + \sqrt{5}}{2} ight) 	imes (1 + \sqrt{5})$$

Expanding the numerator:
$$	ext{Area} = \frac{1}{4} (5 + 5\sqrt{5} + \sqrt{5} + 5)$$
$$	ext{Area} = \frac{10 + 6\sqrt{5}}{4}$$

Simplifying the fraction, we reach the final result:
**$$	ext{Area} = \frac{5 + 3\sqrt{5}}{2}$$**

Animated Solution for Mathematics - Circles: If the tangents drawn at the point $O (0, 0)$ and $P (1 + 5, 2)$ on the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ intersect at the point $Q$ , then the area of the triangle $O P Q$ is equal to

Select Answer:

Visualized Solution

Identify the Circle and Points

The Tangent Formula $T = 0$

Tangent at Origin $O (0, 0)$

Tangent at $P (1 + 5, 2)$

Simplify Tangent at $P$

Finding Intersection Point $Q$

Visualize Triangle $O P Q$

Calculate Base $P Q$

Calculate Height $h$

Final Area Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Power of $T = 0$

The Intersection and the Triangle

Final Calculation

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Select Answer:

Visualized Solution

Identify the Circle and Points

The Tangent Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=0

Tangent at Origin .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } O(0,0)

Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(1+5​,2)

Simplify Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

Visualize Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OPQ

Calculate Base .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Calculate Height .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h