x 
2
 +y 
2
 −20x+10y+100=0, find the equation of the tangents from the origin without simultaneous equstion

To find the equation of the tangents from the origin to the given curve without solving simultaneous equations, we can use the following approach:

### Step 1: Identify the given equation.
The given equation is:
\[
x^2 + y^2 - 20x + 10y + 100 = 0
\]

### Step 2: Convert the equation to a standard circle form.
The equation resembles that of a circle, so we'll complete the square for both \(x\) and \(y\) terms.

1. **For the \(x\) terms**:
   \[
   x^2 - 20x = (x - 10)^2 - 100
   \]
2. **For the \(y\) terms**:
   \[
   y^2 + 10y = (y + 5)^2 - 25
   \]

3. **Substitute back into the original equation**:
   \[
   (x - 10)^2 - 100 + (y + 5)^2 - 25 + 100 = 0
   \]
   Simplifying, we get:
   \[
   (x - 10)^2 + (y + 5)^2 = 25
   \]
   This is the equation of a circle with center at \((10, -5)\) and radius \(5\).

### Step 3: Use the condition for tangency from the origin.
For a line of the form \(y = mx\) to be tangent to the circle \((x - 10)^2 + (y + 5)^2 = 25\), the distance from the center of the circle to the origin must equal the radius.

1. **Distance from the center \((10, -5)\) to the origin**:
   \[
   \text{Distance} = \sqrt{10^2 + (-5)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5}
   \]

2. **Let the line be \(y = mx\)**. The equation of the line can also be written as \(mx - y = 0\).

   The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:
   \[
   d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
   \]
   Here, \(A = m\), \(B = -1\), and \(C = 0\).

   The distance from the origin to the line \(y = mx\) is:
   \[
   d = \frac{|0|}{\sqrt{m^2 + 1}} = \frac{0}{\sqrt{m^2 + 1}} = 0
   \]
   This means we need the line to be at a distance equal to the radius \(5\) from the center.

3. **Equation of the tangents**:
   The line equation tangency condition gives us:
   \[
   \text{Distance} = \frac{|m(10) - 1(-5) + 0|}{\sqrt{m^2 + 1}} = \frac{|10m + 5|}{\sqrt{m^2 + 1}} = 5
   \]
   Solving this, we square both sides:
   \[
   \frac{(10m + 5)^2}{m^2 + 1} = 25
   \]
   \[
   (10m + 5)^2 = 25(m^2 + 1)
   \]
   Expanding both sides:
   \[
   100m^2 + 100m + 25 = 25m^2 + 25
   \]
   Simplify the equation:
   \[
   75m^2 + 100m = 0
   \]
   Factor out:
   \[
   25m(3m + 4) = 0
   \]
   Thus, \(m = 0\) or \(m = -\frac{4}{3}\).

### Step 4: Write the equations of the tangents.
For \(m = 0\), the tangent line is \(y = 0\).

For \(m = -\frac{4}{3}\), the tangent line is:
\[
y = -\frac{4}{3}x
\]

Thus, the equations of the tangents from the origin to the circle are:
\[
y = 0 \quad \text{and} \quad y = -\frac{4}{3}x
\]

Would you like more details on any part of this process, or do you have any other questions?

### Related Questions:
1. How do you complete the square for a general quadratic equation?
2. What is the geometric significance of the tangents to a circle from a given point?
3. How do you derive the distance formula from a point to a line?
4. How do you find the equation of a tangent to a general conic section?
5. What is the difference between the standard form and the general form of a circle's equation?

### Tip:
When solving geometric problems involving circles, always try to convert the given equation to the standard form. This simplifies understanding and solving the problem using known geometric properties.

Explore how to find tangent lines from the origin to a circle given by the equation x^2 + y^2 - 20x + 10y + 100 = 0 without solving simultaneous equations. Learn the key concepts of circle equations, tangent lines, and the distance formula. Suitable for high school students studying circle geometry.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation