If \(3x + y = 0\) is a tangent to the circle with centre at the point \((2, -1)\), then the equation of the other tangent to the circle from the origin, is
Show Hint
- \(x - 3y = 0\)
- \(x + 3y = 0\)
- \(3x - y = 0\)
- \(2x + y = 0\)
The Correct Option is A
Solution and Explanation
To solve the problem, we need to find the equation of another tangent to a circle from the origin. Given that the line \(3x + y = 0\) is a tangent to the circle centered at \((2, -1)\), we can use the property that the perpendicular from the center of the circle to the tangent line will also be the radius of the circle.
The given tangent line to the circle is \(3x + y = 0\). This line has a slope \(m_1 = -3\).
The center of the circle is \((2, -1)\).
Using the point-to-line distance formula, the perpendicular distance (\(d\)) from the center \((x_1, y_1)\) to the line \(Ax + By + C = 0\) is given by:
- \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)
Applying the values to the distance formula:
- \(d = \frac{|3(2) + 1(-1) + 0|}{\sqrt{3^2 + 1^2}} = \frac{|6 - 1|}{\sqrt{9 + 1}} = \frac{5}{\sqrt{10}}\)
The other tangent from the origin will also make the same angle with the origin-center line. The original line is \(3x + y = 0\) having slope \(-3\). Lines from the same point (origin here) having equal angles to the same line will have slopes related as negative reciprocals, so we need a line tangent from origin.
Since the slopes of the tangents are related through negative reciprocals (for perpendicular tangents), the corresponding slope \(m_2\) can either be \(1/3\) or \(-1/3\). However, we want the equation of another tangent \(x - 3y = 0\), which corresponds to the slope being \(1/3\).
Hence, the equation of the other tangent from the origin is \(x - 3y = 0\).
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