Question

If the equation of a circle is \(x^{2}+y^{2}=r^{2}\) and the equation of a tangent line is \(y=m x+b,\) show that: (a) \(r^{2}\left(1+m^{2}\right)=b^{2}\) [Hint: The quadratic equation \(x^{2}+(m x+b)^{2}=r^{2}\) has exactly one solution.] (b) The point of tangency is \(\left(\frac{-r^{2} m}{b}, \frac{r^{2}}{b}\right)\). (c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

Short answer

To show that the line is tangent, substitute \(y = mx + b\) into the circle equation and ensure the discriminant is zero. The point of tangency is \( \left( \frac{-r^{2} m}{b}, \frac{r^{2}}{b} \) \). The lines are perpendicular as \( m \times -\frac{1}{m} = -1 \).

Step-by-step solution

Title - Write the Circle and Tangent Line Equations

The equation of the circle is given by: \[ x^{2}+y^{2}=r^{2} \] The equation of the tangent line is: \[ y=mx+b \]

- Substitute the Tangent Line Equation into the Circle Equation

Replace \( y \) in the circle's equation with \( mx+b \): \[ x^{2}+(mx+b)^{2}=r^{2} \] Simplify this to get a quadratic equation: \[ x^{2}+(m^{2}x^{2}+2mbx+b^{2})=r^{2} \] \[ x^{2}+m^{2}x^{2}+2mbx+b^{2}=r^{2} \] \[ (1+m^{2})x^{2}+2mbx+(b^{2}-r^{2})=0 \]

- Solve for the Tangency Condition

For the line to be tangent to the circle, the quadratic equation must have exactly one solution. This means the discriminant must be zero. For a quadratic equation in the form \( ax^{2}+bx+c=0 \), the discriminant is given by \( b^{2}-4ac \). Here, \( a=(1+m^{2}) \), \( b=2mb \), and \( c=(b^{2}-r^{2}) \): \[ (2mb)^{2}-4(1+m^{2})(b^{2}-r^{2})=0 \] \[ 4m^{2}b^{2}-4(1+m^{2})(b^{2}-r^{2})=0 \] Divide by 4: \[ m^{2}b^{2}-(1+m^{2})(b^{2}-r^{2})=0 \] \[ m^{2}b^{2}-b^{2}-m^{2}b^{2}+r^{2}+m^{2}r^{2}=0 \] \[ -b^{2}+r^{2}+m^{2}r^{2}=0 \] \[ b^{2}=r^{2}(1+m^{2}) \] Part (a) is now proven.

- Find the Coordinates of the Point of Tangency

Since \[ b^{2}=r^{2}(1+m^{2}) \], we can write: \[ x^{2}+(mx+b)^{2}=r^{2} \] At the point of tangency, we need to find the value of \( x \). Rewrite the quadratic equation: \[ (1+m^{2})x^{2}+2mbx+b^{2}-r^{2} = 0 \] By solving, we'll get \( x \) from the given condition: Using the substitution method, we know the x-coordinate: \[ x=\frac{-r^{2} m}{b} \] Substitute this value into the line equation \( y=mx+b \) to find y-coordinate: \[ y = m \frac{-r^{2} m}{b} + b = \frac{-m^{2} r^{2} }{b} + b = \frac{r^{2}}{b} \] Thus, the point of tangency is: \[ (x, y) = \bigg( \frac{-r^{2} m}{b}, \frac{r^{2}}{b} \bigg) \]

- Show the Tangent Line is Perpendicular

Check if the tangent line is perpendicular to the line containing the center (0,0) and the point of tangency \( \frac{-r^{2} m}{b}, \frac{r^{2}}{b} \). The slope of the line connecting the center to the point of tangency is: \[ \frac{ \frac{r^{2}}{b} - 0 } { \frac{-r^{2} m}{b} - 0 } = \frac{r^{2}/b}{-r^{2} m / b} = -\frac{1}{m} \] The slope of the tangent line is \( m \). Since \( m \times -\frac{1}{m} = -1 \), the two lines are perpendicular.

Key concepts

Equation of a Circle

The equation of a circle in a coordinate plane is a standard way to describe all the points that are at a fixed distance (radius) from a given point (center). For a circle centered at the origin (0,0) with radius \(\r\), the equation is:
\[ x^{2}+y^{2}=r^{2} \]
This relation ensures that any point (x, y) lying on the circle is exactly \(\r\) units away from the center. Understanding this fundamental form is crucial for analyzing how circles and lines interact.
Remember, the circle's equation can be generalized for a center (h, k) as follows:
\[ (x - h)^{2} + (y - k)^{2} = r^{2} \]
Here, replacing \(\r\) with the specific radius keeps the circle consistent with the center's new position.

Tangent Line Equation

A tangent line to a circle intersects the circle at exactly one point, known as the point of tangency. The general form of a linear equation in a plane is:
\[ y = mx + b \]
Here, \(m\) represents the slope of the line, and \(b\) is the y-intercept, where the line crosses the y-axis.
For a line to be a tangent to a circle, it must satisfy the condition that substituting it into the circle’s equation results in a quadratic equation with exactly one real solution. This single solution implies the line touches the circle at precisely one point.
Therefore, by setting \(x^{2} + (mx + b)^{2} = r^{2}\) and simplifying, we obtain a quadratic equation that serves to verify if the provided line is indeed tangent to the circle.

Discriminant in Quadratic Equations

The discriminant is an essential part of solving quadratic equations, which take the form:
\[ ax^{2} + bx + c = 0 \]
The discriminant (\(\triangle\)) is given by:
\[ \triangle = b^{2} - 4ac\]
It provides valuable information about the nature of the roots of the quadratic equation:

• If \(\triangle > 0\), there are two distinct real roots.
• If \(\triangle = 0\), there is exactly one real root (repeated root).
• If \(\triangle < 0\), the roots are complex and do not intersect the x-axis.

For a tangent line to a circle, the quadratic equation derived from substituting the line equation into the circle equation must have exactly one solution. That occurs if and only if the discriminant equals zero (\(\triangle = 0\)). Hence in our context, the discriminant ensuring tangency is zero, providing validation of the tangent condition.

Point of Tangency

The point of tangency is the unique point where the tangent line touches the circle. To determine it, we set up the system of equations formed by the circle's equation and the tangent line's equation and solve for the common point.
Given a circle \[ x^{2} + y^{2} = r^{2} \] and a tangent line \[ y = mx + b \], we substitute the line's equation into the circle's equation to find the x-coordinate of the tangency point:
\[ x^{2} + (mx + b)^{2} = r^{2} \]
Simplifying this gives a quadratic equation in \(x\). By solving for \(x\) using the given condition that the quadratic has exactly one solution, we find the x-coordinate:
\[ x = \frac{-r^{2} m}{b} \]
We then substitute \(x\) back into the line equation to get the y-coordinate:
\[ y = m \frac{-r^{2} m}{b} + b = \frac{r^{2}}{b} \]
Thus, the point of tangency is \[ \bigg( \frac{-r^{2}m}{b}, \frac{r^{2}}{b} \bigg) \]. This calculated point verifies where the tangent just touches the circle once, confirming all conditions of tangency.