Question

If \( x-2y=0 \) is a tangent drawn at a point P on the circle \( x^2+y^2-6x+2y+c=0 \), then the distance of the point (6,3) from P is

• A. \( \sqrt{5} \)
• B. \( 2\sqrt{5} \)
• C. \( 4\sqrt{5} \)
• D. \( 5\sqrt{2} \)

Hint:

The normal to a circle at the point of contact always passes through the center. Intersection of tangent and normal gives the point of contact.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We need to find the coordinates of point P (the point of contact). The normal at point P is perpendicular to the tangent and passes through the center of the circle. P is the intersection of the tangent and the normal.
Step 2: Detailed Explanation:

Tangent: \( x - 2y = 0 \implies \) Slope \( m_t = 1/2 \). Circle: \( x^2+y^2-6x+2y+c=0 \). Center \( C(3, -1) \). Normal passes through C and has slope \( m_n = -1/m_t = -2 \). Equation of Normal: \( y - (-1) = -2(x - 3) \) \( y + 1 = -2x + 6 \) \( 2x + y = 5 \). Find P (Intersection of Tangent and Normal): 1. \( x = 2y \) 2. \( 2x + y = 5 \) Substitute (1) in (2): \( 2(2y) + y = 5 \implies 5y = 5 \implies y = 1 \). Then \( x = 2(1) = 2 \). So, \( P(2, 1) \). Distance from \( (6, 3) \) to \( P(2, 1) \): \[ d = \sqrt{(6-2)^2 + (3-1)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]
Step 3: Final Answer:

The distance is \( 2\sqrt{5} \).