Intersection of a Line and a Circle: Fastest Method

Grades 10-11
Student math problem: Intersection of a Line and a Circle: Fastest Method
Original question
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Question: can you teach me the fastest method to find the answer
To find the answer to this problem efficiently, the fastest method is to use the geometric property of circles and lines: the distance from the center of the circle to the line.

1. Identify the Circle Properties

First, we find the center and radius of the circle given by the equation x^2 + y^2 - 12x - 10y + 32 = 0.
  • The center (h, k) is found by taking half the coefficients of x and y and changing their signs: \begin{align*} h &= -\frac{-12}{2} = 6 \\ k &= -\frac{-10}{2} = 5 \end{align*} So, the Center is (6, 5).
  • The Radius r is calculated using r = \sqrt{h^2 + k^2 - \text{constant term}}: \begin{align*} r &= \sqrt{6^2 + 5^2 - 32} \\ r &= \sqrt{36 + 25 - 32} = \sqrt{29} \end{align*}

2. The Intersection Condition

A line Ax + By + C = 0 intersects a circle if the perpendicular distance d from the center of the circle to the line is less than or equal to the radius.
d \le r
If d > r, the line is too far away to touch the circle.

3. Apply the Distance Formula

The perpendicular distance from the center (6, 5) to the line 2x - my + 42 = 0 is: \begin{align*} d = \frac{|2(6) - m(5) + 42|}{\sqrt{2^2 + (-m)^2}} &\le \sqrt{29} \\ \frac{|12 - 5m + 42|}{\sqrt{4 + m^2}} &\le \sqrt{29} \\ \frac{|54 - 5m|}{\sqrt{4 + m^2}} &\le \sqrt{29} \end{align*}

4. Solve the Inequality

Square both sides to eliminate the absolute value and the square root: \begin{align*} \frac{(54 - 5m)^2}{4 + m^2} &\le 29 \\ (54 - 5m)^2 &\le 29(4 + m^2) \\ 2916 - 540m + 25m^2 &\le 116 + 29m^2 \\ 0 &\le 4m^2 + 540m - 2800 \end{align*} Divide by 4: \begin{align*} m^2 + 135m - 700 &\ge 0 \\ (m + 140)(m - 5) &\ge 0 \end{align*}
Since the expression is "greater than or equal to zero," m must lie outside the roots -140 and 5. Therefore, m \le -140 or m \ge 5.
The correct answer is D.

Visualizing the Geometry

In the diagram below, you can see the Circle (in blue) and the Line (in red). The center is marked at (6, 5). As you adjust the slider for m, notice that the line only touches or enters the circle when m reaches the "critical" boundary values of 5 or -140. Outside this range, the distance d from the center to the line becomes greater than the radius, and they no longer intersect.
coordinate-geometrycirclesalgebra