The tangent to the circle centered at (0,0) with radius 1 at point \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) is given by:
\( x + y = \sqrt{2} \)
Understanding the Tangent to a Circle
The problem asks for the equation of the tangent line to a circle at a specific point. The circle is centered at the origin (0,0) and has a radius of 1. The point where the tangent touches the circle is given as \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \).
Circle Equation and Point Verification
The standard equation of a circle centered at \( (h, k) \) with radius \( r \) is \( (x-h)^2 + (y-k)^2 = r^2 \). For a circle centered at the origin \( (0,0) \) with radius \( r=1 \), the equation is:
\( x^2 + y^2 = 1^2 \)
\( x^2 + y^2 = 1 \)
Before finding the tangent, it's good practice to check if the given point \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) actually lies on the circle. Substitute the coordinates into the circle's equation:
\( \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} + \frac{1}{2} = 1 \)
Since the equation holds true, the point \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) is indeed on the circle.
Formula for Tangent at a Point on the Circle
For a circle with the equation \( x^2 + y^2 = r^2 \), the equation of the tangent line at a point \( (x_0, y_0) \) on the circle is given by the formula:
\( xx_0 + yy_0 = r^2 \)
Calculating the Tangent Equation
In this problem, the circle is \( x^2 + y^2 = 1 \), so \( r^2 = 1 \). The point of tangency is \( (x_0, y_0) = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \). Using the formula \( xx_0 + yy_0 = r^2 \), we substitute the values:
\( x \left(\frac{1}{\sqrt{2}}\right) + y \left(\frac{1}{\sqrt{2}}\right) = 1 \)
To simplify the equation, we can multiply the entire equation by \( \sqrt{2} \):
\( \sqrt{2} \left( x \frac{1}{\sqrt{2}} + y \frac{1}{\sqrt{2}} \right) = \sqrt{2} \times 1 \)
\( x + y = \sqrt{2} \)
Comparing with Options
The calculated equation of the tangent line is \( x + y = \sqrt{2} \). Let's compare this with the given options:
- Option 1: \( x - y = 0 \)
- Option 2: \( x + y = \sqrt{2} \)
- Option 3: \( 2\sqrt{2}x - 3\sqrt{2}y = -1 \)
- Option 4: \( 3\sqrt{2}x + \sqrt{2}y = 4 \)
Our derived equation \( x + y = \sqrt{2} \) matches Option 2.
| Step | Description | Formula/Calculation |
|---|---|---|
| 1 | Identify circle equation \(x^2+y^2=r^2\) | \(x^2+y^2=1\) (so \(r^2=1\)) |
| 2 | Identify point of tangency \((x_0, y_0)\) | \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) |
| 3 | Apply tangent formula \(xx_0 + yy_0 = r^2\) | \(x(\frac{1}{\sqrt{2}}) + y(\frac{1}{\sqrt{2}}) = 1\) |
| 4 | Simplify the equation | \(x + y = \sqrt{2}\) |
Revision Table: Key Concepts for Tangent Lines
| Concept | Description | Formula (for circle \(x^2+y^2=r^2\)) |
|---|---|---|
| Circle Equation (Center Origin) | Equation for a circle centered at (0,0) with radius r. | \(x^2 + y^2 = r^2\) |
| Point on Circle | A point \((x_0, y_0)\) satisfies the circle's equation. | \(x_0^2 + y_0^2 = r^2\) |
| Tangent at Point \((x_0, y_0)\) | The equation of the line tangent to the circle at the point \((x_0, y_0)\) on its circumference. | \(xx_0 + yy_0 = r^2\) |
Additional Information on Circle Tangents
Finding the equation of a tangent line is a fundamental concept in coordinate geometry. Here are some related points:
- General Circle Equation: If the circle is centered at \( (h, k) \) with radius \( r \), its equation is \( (x-h)^2 + (y-k)^2 = r^2 \). The tangent at \( (x_0, y_0) \) is \( (x-h)(x_0-h) + (y-k)(y_0-k) = r^2 \). Notice that the formula \( xx_0 + yy_0 = r^2 \) is a special case of this when \( (h, k) = (0,0) \).
- Radius Perpendicular to Tangent: An important property is that the radius drawn from the center of the circle to the point of tangency is always perpendicular to the tangent line at that point. We could also use this property (finding the slope of the radius and then the negative reciprocal for the tangent's slope) to find the tangent equation, especially if the point of tangency is given.
- Calculus Method: The slope of the tangent line can also be found using calculus by implicitly differentiating the circle's equation \( x^2 + y^2 = r^2 \) with respect to \( x \) to find \( \frac{dy}{dx} \), which gives the slope at any point \( (x, y) \).
Understanding these different methods helps in tackling various types of tangent problems involving circles.
Similar Questions
If the distance of the point (4,6,8) from the plane \( \vec{r} \cdot (6\hat{i} - 12\hat{j} + 4\hat{k}) = a \) is 1, then \( a \) is:
If the equation of a line \( PQ \) is:
\[ \frac{x+1}{2} = \frac{2-y}{5} = \frac{z+6}{7} \]
then the direction cosines of a line parallel to \( PQ \) are:
Find the equation of a line through the point (-2, 1, 3) and parallel to the line:
\[ \frac{x - 2}{4} = \frac{y + 3}{-3}, \quad z = -2 \]
Important Questions from Three-Dimensional Geometry
There are two bags. Bag-1 contains 4 white and 6 black balls and Bag-2 contains 5 white and 5 black balls.
A die is rolled. If it shows a number divisible by 3, a ball is drawn from Bag-1; otherwise, a ball is drawn from Bag-2.
If the ball drawn is not black in colour, the probability that it was not drawn from Bag-2 is:
The unit vector perpendicular to each of the vectors $ \vec{a} + \vec{b}$ and $ \vec{a} - \vec{b}$, where, $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ is :
The direction cosines of the line which is perpendicular to the lines with direction ratios (1, -2, -2) and (0, 2, 1) are:
If the distance of the point (4,6,8) from the plane \( \vec{r} \cdot (6\hat{i} - 12\hat{j} + 4\hat{k}) = a \) is 1, then \( a \) is:
If the equation of a line \( PQ \) is:
\[ \frac{x+1}{2} = \frac{2-y}{5} = \frac{z+6}{7} \]
then the direction cosines of a line parallel to \( PQ \) are:
