The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle

Q: Wait, why does the normal always pass through the center? Can't it be any line perpendicular to the tangent?

That is a fantastic question! You are absolutely right that the normal is perpendicular to the tangent. However, in a circle, the radius at the point of tangency is always perpendicular to the tangent line. Since the radius connects the center to the point of tangency, the line containing the radius is the normal. Therefore, the normal must pass through the center.

Q: Why can't we just use the slope-point form for the tangent instead of the $xx_1 + yy_1 = r^2$ formula?

You absolutely can! The formula $xx_1 + yy_1 = r^2$ is just a shortcut derived from the general equation of a tangent. If you prefer, you can find the slope of the radius $OP$, which is $\sqrt{3}$, then find the slope of the tangent (the negative reciprocal, $-1/\sqrt{3}$), and use the point-slope form $y - y_1 = m(x - x_1)$. Both methods will lead you to the exact same equation, $x + \sqrt{3}y = 4$.

Q: Is the triangle $OAP$ always right-angled?

In this specific case, the triangle $OAP$ is not necessarily a right-angled triangle. The angle at $P$ is $90^\circ$ because the tangent and normal are perpendicular, but the triangle is formed by the origin, the x-intercept of the tangent, and the point $P$. You should calculate the area using the base-height method rather than assuming it is a right triangle.

Question

The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle $x^2 + y^2 = 4$ at $(1, rac{\sqrt{3}}{1})$ is .........

Sigma Percentile · Accepted Answer

### The Geometry of Perfection: Unveiling the Triangle

Welcome, fellow explorer of mathematics! Today, we are not just solving a coordinate geometry problem; we are embarking on a journey to uncover the hidden relationships between lines and curves.

Imagine you are standing on a coordinate plane, looking at a perfect circle defined by the equation $x^2 + y^2 = 4$. This is a circle of radius $r = 2$, centered gracefully at the origin $O(0,0)$.

Our mission is to find the area of a triangle formed by the positive x-axis, the tangent to this circle, and the normal to this circle, all meeting at a specific point $P(1, \sqrt{3})$. Let us peel back the layers of this problem together.

### Phase 1: The Point of Contact

Before we dive into the algebra, we must anchor ourselves. We are given the point $P(1, \sqrt{3})$.

Is it truly on our circle? Let us test it. Substituting $x = 1$ and $y = \sqrt{3}$ into $x^2 + y^2$, we get:

$$1^2 + (\sqrt{3})^2 = 1 + 3 = 4$$

It matches perfectly! Point $P$ is indeed a guardian of the circle's boundary. Knowing this, we can now confidently proceed to find the lines that define our triangle.

### Phase 2: The Tangent Line

To find the tangent at $P(1, \sqrt{3})$, we could use calculus, but why take the long road when we have a beautiful shortcut? For any circle $x^2 + y^2 = r^2$, the tangent at $(x_1, y_1)$ is given by the elegant equation $xx_1 + yy_1 = r^2$.

Substituting our values, $x_1 = 1$, $y_1 = \sqrt{3}$, and $r^2 = 4$, we get:

$$x(1) + y(\sqrt{3}) = 4 \Rightarrow x + \sqrt{3}y = 4$$

This line is the first side of our triangle. To find where it meets the x-axis, we set $y = 0$, yielding $x = 4$. Thus, our tangent intersects the x-axis at point $A(4, 0)$.

### Phase 3: The Normal Line

Now, let us consider the normal. There is a profound geometric truth here: the normal to a circle at any point always passes through the center.

Since our center is the origin $O(0,0)$, the normal is simply the line passing through $O(0,0)$ and $P(1, \sqrt{3})$. The slope of this line is:

$$m = \frac{\sqrt{3} - 0}{1 - 0} = \sqrt{3}$$

Therefore, the equation of the normal is $y = \sqrt{3}x$. Where does this normal meet the x-axis? Setting $y = 0$ gives $x = 0$. So, the normal intersects the x-axis right at the origin $O(0,0)$.

### Phase 4: The Final Calculation

We have our vertices: $O(0,0)$, $A(4,0)$, and $P(1, \sqrt{3})$. We are looking for the area of $\Delta OAP$.

The base of this triangle lies on the x-axis, stretching from the origin $O$ to the point $A(4,0)$. The length of this base is $4 - 0 = 4$ units.

The height of the triangle is the perpendicular distance from $P$ to the x-axis, which is simply the y-coordinate of $P$, namely $\sqrt{3}$. Using the classic area formula, $	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$, we calculate:

$$	ext{Area} = \frac{1}{2} 	imes 4 	imes \sqrt{3} = 2\sqrt{3}$$

### Conclusion

And there it is! The area is **$2\sqrt{3}$** square units. It is a beautiful result, isn't it?

By understanding the geometric properties of the circle—that the normal passes through the center and the tangent can be found with a simple formula—we transformed a potentially daunting problem into a clear, logical path. Keep this clarity with you as you tackle more complex problems. You are doing great!

Animated Solution for Mathematics - Circles: The area of the triangle formed by the positive x-axis and the normal and the tangent to the circle $x^{2} + y^{2} = 4$ at $(1, rac{\sqrt{3}}{1})$ is .........

Visualized Solution

Visualizing the Circle $x^{2} + y^{2} = 4$

Identifying Point $P (1, 3)$

Formula for the Tangent Line

Finding the Tangent Equation

Finding Point $A (4, 0)$

Understanding the Normal to a Circle

Finding the Normal Equation

Finding the X-intercept of the Normal

Identifying the Triangle $O A P$

Calculating the Area of $Δ O A P$

Key Takeaways

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Perfection

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

The tangent and the normal lines at the point $(3, 1)$ to the circle $x^{2} + y^{2} = 4$ and the x-axis form a triangle. The area of this triangle (in square units) is :

Let the tangents at two points $A$ and $B$ on the circle $x^{2} + y^{2} - 4 x + 3 = 0$ meet at origin $O (0, 0)$ . Then the area of the triangle of $O A B$ is

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

The area (in sq. units) of the smaller of the two circles that touch the parabola, $y^{2} = 4 x$ at the point $(1, 2)$ and the x-axis is :-

If the area of the triangle formed by the positive x -axis, the normal and the tangent to the circle $(x - 2)^{2} + (y - 3)^{2} = 25$ at the point (5,7) is A, then 24 A is equal to Note: NTA has dropped this question in the final official answer key.

Let the tangent to the parabola $S : y^{2} = 2 x$ at the point $P (2, 2)$ meet the $x$ -axis at $Q$ and normal at it meet the parabola $S$ at the point $R$ . Then the area (in sq. units) of the triangle $P QR$ is equal to:

Let the tangent to the circle $x^{2} + y^{2} = 25$ at the point R(3, 4) meet x -axis and y -axis at point P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then $r^{2}$ is equal to :

The minimum area of triangle formed by the tangent to the $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ & coordinate axes is