To solve this problem, we need to find another tangent from the origin to the circle for which \(3x + y = 0\) is already a tangent. The center of the circle is given as \((2, -1)\), and we will use the concept of tangents to a circle from an external point.
Let's go through the solution step-by-step:
- First, identify the equation of the circle.
- For a line \(3x + y = 0\) to be tangent to a circle with center at \((2, -1)\), the perpendicular distance from the center to the line should be equal to the radius \(r\) of the circle.
- Compute the perpendicular distance from the point \((2, -1)\) to the line \(3x + y = 0\).
- The perpendicular 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, \(A = 3\), \(B = 1\), and \(C = 0\), and the point is \((2, -1)\).
- So, the distance is: \(d = \frac{|3(2) + 1(-1) + 0|}{\sqrt{3^2 + 1^2}} = \frac{5}{\sqrt{10}}\)
- Since this distance represents the radius \(r\) of the circle, we now find the equation of the circle as:
Center: \((2, -1)\) and radius \(r = \frac{5}{\sqrt{10}}\).
Equation of the circle: \((x - 2)^2 + (y + 1)^2 = \left(\frac{5}{\sqrt{10}}\right)^2\)
- Find the equation of the other tangent from the origin.
- Tangents from an external point implies there are two tangents from the origin to the circle.
- Since \(3x + y = 0\) is already one tangent, we check the other lines.
- The other tangent must satisfy the condition of orthogonal lines to \(3x + y = 0\), thus its slope will be the negative reciprocal. Slope of \(3x + y = 0\) is \(-3\), the perpendicular slope is \(\frac{1}{3}\).
- The equation of the line from origin having slope \(\frac{1}{3}\) is of the form \(y = \frac{1}{3}x\) which gives us \(x - 3y = 0\).
Hence, the equation of the other tangent to the circle from the origin is \(x - 3y = 0\).