Question

If the equation of one tangent to the circle with center at $(2, -1)$ from the origin is $3x + y = 0$, then the equation of the other tangent through the origin is

• A. 3x-y=0
• B. x+3y=0
• C. x-3y=0
• D. x+2y=0

Answer 1

The Correct Option is C

The general equation of a circle with center \( (h, k) \) and radius \( r \) is: \[ (x - h)^2 + (y - k)^2 = r^2 \] For this circle, the center is \( (2, -1) \), so the equation of the circle becomes: \[ (x - 2)^2 + (y + 1)^2 = r^2 \]

We are given the equation of one of the tangents from the origin, which is: \[ 3x + y = 0 \] This equation represents a straight line with slope \( -3 \) (since it can be written as \( y = -3x \)).

Step 1: Find the slope of the radius at the point of tangency
The radius of the circle is always perpendicular to the tangent at the point of contact. Therefore, the product of the slopes of the radius and the tangent is \( -1 \).
If the slope of the tangent is \( -3 \), the slope of the radius is: \[ \text{slope of radius} = \frac{1}{3} \]

Step 2: Find the equation of the radius
The radius passes through the center of the circle, which is \( (2, -1) \), and has a slope of \( \frac{1}{3} \). The equation of the radius is: \[ y - (-1) = \frac{1}{3}(x - 2) \] Simplifying: \[ y + 1 = \frac{1}{3}(x - 2) \Rightarrow y = \frac{1}{3}x - \frac{5}{3} \]

Step 3: Find the equation of the second tangent
The second tangent will have a slope that is symmetric to the first tangent with respect to the radius. Since the first tangent has slope \( -3 \), the second tangent will have slope \( 3 \). The equation of the second tangent passing through the origin with slope \( 3 \) is: \[ y = 3x \] Rearranging: \[ x - 3y = 0 \]

Answer:

\[ \boxed{x - 3y = 0} \]