Question

For the circle \( x-2=5\cos\theta \), \( y+1=5\sin\theta \) where \( \theta \) is the parameter, the line \( x=1+\frac{\sqrt{3}}{2}r \), \( y=-2+\frac{r}{2} \) where \( r \) is the parameter, is a

• A. Chord of the circle other than diameter
• B. Tangent of the circle
• C. Diameter of the circle
• D. Line that does not meet the circle

Hint:

Calculating the "Power of the Point" ($S_1$) is the quickest way to determine position. Negative = Inside (Chord), Zero = On circle (Tangent if direction matches, chord otherwise), Positive = Outside (Secant, Tangent, or Non-intersecting).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:

Identify the circle's center and radius, and a fixed point on the line. Determine the position of the point relative to the circle. Check if the line passes through the center.
Step 2: Detailed Explanation:

Circle Analysis:
\( x - 2 = 5\cos\theta, \ y + 1 = 5\sin\theta \). Squaring and adding: \( (x-2)^2 + (y+1)^2 = 25 \). Center \( C(2, -1) \), Radius \( R = 5 \). Line Analysis:
Given \( x = 1 + \frac{\sqrt{3}}{2}r, \ y = -2 + \frac{r}{2} \). This represents a line passing through point \( P(1, -2) \) with a specific direction. Check position of P relative to circle: \( S_1 = (1-2)^2 + (-2+1)^2 - 25 = 1 + 1 - 25 = -23 \). Since \( S_1 \textless 0 \), point P lies inside the circle. Any line passing through an interior point intersects the circle at two points, making it a chord. Check for Diameter:
Does the center \( (2, -1) \) lie on the line? Substitute \( x=2 \) into line eq: \( 2 = 1 + \frac{\sqrt{3}}{2}r \implies r = \frac{2}{\sqrt{3}} \). Substitute \( r = \frac{2}{\sqrt{3}} \) into y-equation: \( y = -2 + \frac{1}{2}\left(\frac{2}{\sqrt{3}}\right) = -2 + \frac{1}{\sqrt{3}} \approx -2 + 0.57 = -1.43 \). But the center's y-coordinate is -1. Since \( -1 \neq -1.43 \), the center is not on the line. Thus, it is a chord but not a diameter.
Step 3: Final Answer:

It is a chord of the circle other than diameter.