The area of the triangle formed by the tangents from the point (4, 3) to the the circle x² + y² = 9

Q: Wait, why is the line joining the center to the external point perpendicular to the chord of contact?

This is a beautiful property of circle geometry! The two tangents drawn from an external point to a circle are equal in length. This means the triangle formed by the external point and the two points of contact is isosceles. The line segment from the center to the external point is the angle bisector of the angle between the tangents, and by symmetry, it must be the perpendicular bisector of the chord of contact.

Q: Can I just find the coordinates of the points of contact $T$ and $T'$ and use the distance formula?

You absolutely can! You would solve the system of equations formed by the circle $x^2 + y^2 = 9$ and the chord of contact $4x + 3y = 9$. While this is a valid algebraic approach, it involves solving a quadratic equation and dealing with potentially messy coordinates. The geometric method we used is much faster and less prone to calculation errors.

Q: Why is the height of the triangle $PR = OP - OR$?

Great observation! The chord of contact is perpendicular to the line segment $OP$. Let $R$ be the point where the chord intersects $OP$. Since $O$, $R$, and $P$ are collinear in that order, the distance from the external point $P$ to the chord (which is the height of our triangle) is simply the total distance $OP$ minus the distance from the origin to the chord, $OR$.

Question

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 .........

Sigma Percentile · Accepted Answer

### The Geometry of Tangents: A Journey to the Chord of Contact

Imagine you are standing at the origin of a coordinate plane, looking at a circle with a radius of $3$. It is a perfect, symmetrical shape defined by $x^2 + y^2 = 9$.

Now, place a point $P$ at coordinates $(4, 3)$. As you look from $P$ toward the circle, you can draw two lines that just graze the edge of the circle—these are your tangents.

These two lines, along with the chord that connects the two points where they touch the circle, form a triangle. Our mission is to calculate the area of this triangle.

### Phase 1: The Chord of Contact

First, we must define the base of our triangle. This base is the chord of contact, the line segment joining the two points of tangency, let's call them $T$ and $T'$.

There is a powerful, elegant formula for this: for any external point $(x_1, y_1)$ and a circle $x^2 + y^2 = R^2$, the equation of the chord of contact is $xx_1 + yy_1 = R^2$.

Substituting our values, $x_1 = 4$, $y_1 = 3$, and $R^2 = 9$, we get the equation:

$$4x + 3y = 9$$

This line is the foundation of our triangle.

### Phase 2: The Geometry of the Altitude

Now, let's look at the height. The line segment connecting the origin $O(0,0)$ to our external point $P(4,3)$ is the axis of symmetry for this entire setup. It is perpendicular to our chord of contact.

Let $R$ be the intersection point of $OP$ and the chord. The distance $OP$ is easily found using the distance formula:

$$OP = \sqrt{4^2 + 3^2} = 5$$

Next, we need the distance $OR$, which is the perpendicular distance from the origin to the line $4x + 3y - 9 = 0$. Using the standard formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$, we find:

$$OR = \frac{|-9|}{\sqrt{4^2 + 3^2}} = \frac{9}{5} = 1.8$$

The height of our triangle is the segment $PR$, which is simply:

$$PR = OP - OR = 5 - 1.8 = 3.2$$

### Phase 3: The Base and the Final Area

We are almost there! To find the base $TT'$, we first look at the right-angled triangle $	riangle OTR$, where $OT$ is the radius of the circle ($3$ units).

By the Pythagorean theorem, the half-base $TR$ is:

$$TR = \sqrt{OT^2 - OR^2} = \sqrt{3^2 - 1.8^2} = \sqrt{9 - 3.24} = \sqrt{5.76} = 2.4$$

Since $OP$ bisects the chord, the total base $TT'$ is $2 	imes 2.4 = 4.8$.

Finally, the area of $	riangle PTT'$ is calculated as:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{Base} 	imes 	ext{Height} = \frac{1}{2} 	imes 4.8 	imes 3.2 = 7.68$$

And there it is—the elegance of geometry reveals the answer to be **$7.68$ square units**. You have successfully navigated the relationship between tangents, chords, and the symmetry of the circle!

Animated Solution for Mathematics - Circles: 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 .........

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Equation of Chord of Contact $T T^{'}$

Distance from Origin to Point $P$

Perpendicular Distance $O R$

Height of the Triangle $P R$

Half-Base Length $T R$

Total Base Length $T T^{'}$

Final Area Calculation

The Sigma Insight: Length of Tangent and Chord of Contact

The Geometry of Tangents

Phase 1

Phase 2

Phase 3

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Equation of Chord of Contact .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TT′

Distance from Origin to Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Perpendicular Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OR

Height of the Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PR

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

Total Base Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TT′