Verified Answer :
- Identify the center \((-5, -3)\) and radius \(\sqrt{80} = 4\sqrt{5}\) of the circle.
- Define the line equation as \(2x - y + c = 0\) based on the given gradient of 2.
- Use the perpendicular distance formula to relate the distance from the center to the line to the circle's radius.
- Solve the resulting absolute value equation \(|c - 7| = 20\) to find \(c = 27\) and \(c = -13\).
- State the final equations: \(\boxed{y = 2x + 27}\) and \(\boxed{y = 2x - 13}\).
Explanation
- Analyze the circle equation.
The circle is given by the equation \((x+5)^2 + (y+3)^2 = 80\). By comparing this to the standard form of a circle's equation \((x-h)^2 + (y-k)^2 = r^2\), we identify the center \((h, k)\) and the radius \(r\):
- Center \((h, k) = (-5, -3)\)
- Radius squared \(r^2 = 80\), so the radius \(r = \sqrt{80} = 4\sqrt{5}\).
- Set up the distance equation for tangency.
A line with gradient \(m=2\) can be written in the form \(y = 2x + c\), which can be rearranged to the general form \(2x - y + c = 0\). For this line to be a tangent to the circle, its perpendicular distance \(d\) from the center of the circle \((-5, -3)\) must be equal to the radius \(r = 4\sqrt{5}\).
The perpendicular distance formula from a point \((x_0, y_0)\) to a line \(Ax + By + C_0 = 0\) is:
\[d = \frac{|Ax_0 + By_0 + C_0|}{\sqrt{A^2 + B^2}}\]
Substituting the values \(A=2\), \(B=-1\), \(C_0=c\), \(x_0=-5\), and \(y_0=-3\):
\[d = \frac{|2(-5) - 1(-3) + c|}{\sqrt{2^2 + (-1)^2}} = \frac{|-10 + 3 + c|}{\sqrt{4+1}} = \frac{|c - 7|}{\sqrt{5}}\]
- Solve for the constant c and state the final equations.
We equate the perpendicular distance \(d\) to the radius \(r\):
\[\frac{|c - 7|}{\sqrt{5}} = 4\sqrt{5}\]
Multiplying both sides by \(\sqrt{5}\):
\[|c - 7| = 4\sqrt{5} \times \sqrt{5} = 4 \times 5 = 20\]
This absolute value equation gives us two possible cases for \(c\):
Case 1: \(c - 7 = 20 \implies c = 27\)
Case 2: \(c - 7 = -20 \implies c = -13\)
Thus, the two possible equations for the line \(l\) are \(y = 2x + 27\) and \(y = 2x - 13\).
Examples
In civil engineering, determining the tangency of a road path (represented as a line) to a circular roundabout is a common task. If a planner models a roundabout's boundary with a circle equation and defines a road segment with a specific gradient, solving for the intercept allows them to calculate the exact points where the road meets the roundabout boundary, ensuring the traffic flow is smooth and the intersection design is geometrically accurate. For example, if a circular feature is defined as \((x+5)^2 + (y+3)^2 = 80\), the possible roads with gradient 2 that touch the edge of this feature are given by \(y = 2x + 27\) and \(y = 2x - 13\).